sample problem solving in acceleration

v = 0 m/s

v = 88.3 m/s

( 112 m/s) 2 = (0 m/s) 2 + 2*(a)*(398 m)

12544 m 2 /s 2 = 0 m 2 /s 2 + (796 m)*a

12544 m 2 /s 2 = (796 m)*a

(12544 m 2 /s 2 )/(796 m) = a

a = 15.8 m/s 2

Return to Problem 19

v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(91.5 m)

0 m 2 /s 2 = v i 2 - 1793 m 2 /s 2

1793 m 2 /s 2 = v i 2

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr

Return to Problem 20

Newton’s second law of motion – problems and solutions

Solved problems in Newton’s laws of motion – Newton’s second law of motion 

Wanted : net force (∑F)

∑ F = (1 kg)(5 m/s 2 ) = 5 kg m/s 2 = 5 Newton

2. Mass of an object = 1 kg, net force ∑F = 2 Newton. Determine the magnitude and direction of the object’s acceleration….

Mass (m) = 1 kg

The direction of the acceleration = the direction of the net force (∑F)

Mass (m) = 2 kg

∑ F = F 1 – F 2 = 5 – 3 = 2 Newton

F 1x = F 1 cos 60 o = (10)(0.5) = 5 Newton

Wanted : The magnitude and direction of the acceleration (a)

5. F 1 = 10 Newton, F 2 = 1 Newton, m 1 = 1 kg, m 2 = 2 kg. The magnitude and direction of the acceleration is…

A 40-kg block accelerated by a force of 200 N. Acceleration of the block is 3 m/ s 2 . Determine the magnitude of friction force experienced by the block.

Force (F) = 200 N

∑ F = net force, m = mass, a = acceleration

200 – F g = 120

7. Block A with a mass of 100-gram place above block B with a mass of 300 gram, and then block b pushed with a force of 5 N vertically upward. Determine the normal force exerted by block B on block A.

Mass of block B (m B ) = 300 gram = 0.3 kg

Weight of block A (w A ) = (0.1 kg)(10 m/s 2 ) = 1 kg m/s 2 = 1 Newton

N A = normal force exerted by block B on block A (Act on block A)

Apply Newton’s second law of motion on both blocks :

N A and N A ’ are action-reaction forces that have the same magnitude but opposite in direction so eliminated from the equation.

5 – 4 = (0.4) a

N A – w A = m A a

8. An object with weight of 4 N supported by a cord and pulley. A force of 2 N acts on the block and one end of the cord pulled by a force of 9 N. Determine the net force acts on object X.

Weight of X (w X ) = 4 Newton

The tension force has the same magnitude in all part of the cord. So the tension force is 9 N.

The net force act on the object X is 3 Newton, vertically upward.

The correct answer is A.

10. Tom and Andrew push an object on the smooth floor. Tom push the object with a force of 5.70 N. If the mass of the object is 2.00 kg and acceleration experienced by the object is 2.00 ms -2 , then determine the magnitude and direction of force act by Tom.

C. 2.30 N and its direction is opposite with force acted by Andrew.

Wanted : Magnitude and direction of force acted by Tom (F 2 ) ?

ΣF = 4 N + 2 N – 3 N = 6 N – 3 N = 3 Newton, leftward

ΣF = 3 N + 4 N + 2 N = 9 Newton, rightward

Net force (ΣF) = 25 N + 30 N – 15 N = 40 N

(2) B ooks on paper are not falling when the paper is pulled quickly

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Animated Physics Lessons

F=ma Practice Problems

F=ma problem set.

Practice solving for net force, using Newtons second law (F=ma), and relating F=ma to the acceleration equations.

In these practice problems we will either use F=ma or our 1D motion acceleration equations to solve force problems.

1. What is the acceleration of the 15 kg box that has 500 N of force applied to the right?

a = 33.33 m/s 2 Right

2. What is the acceleration of the 25 kg box that has 50 N of force applied to the right?

                a=2.0 m/s 2 Right

3. What is the acceleration of the 3 kg box that has 25 N of force applied to the right and 55 N left?

               a = 10.0 m/s 2 Left

4. What is the acceleration of the 5 kg box that has a 25 N force and 50 N force applied both right?

            a = 15.0 m/s 2 Right

5. What is the acceleration of the 25 kg box that has a 100 N force north and 50 N force east applied?

             a= 4.47 m/s 2

5b. What direction would this box accelerate?

            63.43° North of East

6. Does a 795 kg Lorinser speedy, 6300 kg elephant, or 8.6 kg wagon have more inertia and why?

            6300 kg Elephant

            The more mass the more inertia

7. How much force is required to accelerate a 795 kg Lorinser Speedy by 15 m/s 2 ?

            F = 11925 N

8. How much force is required to accelerate an 8.6 kg wagon by 15 m/s 2 ?

            F = 129 N

9. How much does a 6300 kg elephant accelerate when you apply 500 N of force?

            a = 0.0794 m/s 2

10. What is the mass of an object if it takes a net force of 40 N to accelerate at a rate of 0.88 m/s 2 ?

            m = 45.45 kg

11. How much force is required to accelerate a 0.142 kg baseball to 44.7 m/s during a pitchers 1.5 meter delivery?

            F = 94.58 N

12. A 0.050 kg golf ball leaves the tee at a speed of 75.0 m/s. The club is in contact with the ball for 0.020 s. What is the net force of the club on the ball?

                F = 187.5 N

13. A 90.0 kg astronaut receives a 30.0 N force from her jetpack. How much faster is she be moving after 2.00 seconds?

            0.667 m/s faster

14. A 795 kg car starts from rest and travels 41 m in 3.0 s. How much force did the car engine provide?

            F = 7242 N

15. Joe and his sailboat have a combined weight of 450 kg. How far has Joe sailed when he started at 5 m/s and a gust of wind provided 600 Newtons of force for 4 seconds?

            x = 30.64 m

16. Tom pulls a 45 kilogram wagon with a force of 200 Newtons at a 15° angle to the horizontal from rest. How much faster will the wagon be moving after 2 seconds?

            v f = 8.58 m/s

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Acceleration

Suppose an object moves from one point to another such that its velocity at the initial point is different from that at the final point. Acceleration is defined as the rate at which the velocity changes. It is a vector quantity having both magnitude and direction.

To understand acceleration, let us take a numerical example. An object starts from rest and picks up speed such that its velocity becomes 5 m/s in 10 seconds. It took 10 s to go from 0 m/s to 5 m/s. Then, the rate at which its velocity changed is: (5 m/s – 0)/10 s = 0.5 m/s 2 . Therefore, its acceleration is 0.5 m/s 2 .

• A plane taking off
• Pressing the gas pedal of a car
• A roller coaster riding on its tracks
• A fruit falling from a tree
• Throwing a football
• Going up and down an elevator

Average Acceleration

Acceleration is the rate of change of velocity. In other words, it is the change in velocity over a given time interval. Suppose v i and v f are the initial and final velocities of the object at time t i and t f , respectively. Then, the average acceleration is

Δv: Change in velocity

Δt: Time interval over which the change occurred

SI Unit : meters per second squared or m/s 2

Imperial unit : feet per second squared or ft/s

• Acceleration and velocity must not be confused. Velocity is the change in position with respect to time. If the object moves with a constant velocity, then Δv = 0, and its acceleration is zero.
• In the above equation, acceleration is the change in velocity over the change in time. It does not mean that if the velocity is high, the acceleration is high, or if the velocity is low, the acceleration is low. The velocity can be high, the acceleration can still be low, and vice versa.
• If the velocity changes by equal amounts in equal time intervals, the acceleration will remain constant. It is known as uniform or constant acceleration. The graphs below illustrate the difference between constant acceleration, uniformly increasing acceleration, and nonuniformly increasing acceleration.
• Can acceleration be negative . Generally, acceleration refers to situations where the velocity increases with time. In this case, the acceleration is a positive quantity. However, if the velocity decreases with time, the magnitude of acceleration will be negative. The object slows down, and the phenomenon is called deceleration or retardation.

Instantaneous Acceleration

The average acceleration is measured over a long interval of time. On the other hand, instantaneous acceleration is the acceleration calculated for an infinitesimally small time interval. It can be found by setting the limit of the time interval in the above equation to zero.

Therefore, acceleration is the first derivative of the velocity with respect to time. Note that the velocity is the derivative of distance (x) over time. We can rewrite the above equation as

Therefore, acceleration is the second derivative of distance with respect to time.

Acceleration from Newton’s Second Law

Acceleration can also be calculated directly from Newton’s Second Law . According to this law, the net force (F) on an object is given by the product of its mass (m) and acceleration (a). Mathematically, it is represented by

Rearranging this equation gives the acceleration a.

Direction of Acceleration

The direction of acceleration depends on whether the object is speeding or slowing down and the direction of travel. Imagine a car traveling on a straight road such that it is speeding up. Its velocity increases with time. Its direction is along the direction of travel. Now, suppose the car is slowing down. Its velocity decreases with time. Its direction is opposite to the direction of travel. This type of acceleration, in which the acceleration vector is along the displacement vector, is called linear acceleration .

Suppose the car makes a turn around a curve path. Its velocity vector will change. We can use the concept of vector resolution to study its acceleration. Its velocity vector can be resolved into two components. One component is directed toward the center of curvature of the curve path, called the radial velocity. Another component is perpendicular to the radial velocity, called the tangential velocity. The rate of change of radial velocity is called radial acceleration and is directed radially inward. The rate of change of tangential velocity is called tangential acceleration . It is directed tangentially to the point under consideration on the curve path, as shown in the image below.

Suppose the car moves around a circular path. In that case, its radial acceleration is called centripetal acceleration , discussed in our article on centrifugal force .

Example Problems

Problem 1 : A motorcyclist starts from rest and speeds up uniformly to 60 miles per hour (mph) in 12 seconds. What was the magnitude of the average acceleration of the motorcyclist?

Given v i = 0, v f = 60 mph, Δt = 12 s

Let us convert mph to m/s.

60 mph = (60 miles x 1.6 km/mile x 1000 m/km)/(1 hr x 3600 s/hr) = 26.67 m/s

The acceleration is

a = (v f – v i )/Δt = (26.67 m/s – 0)/12 s = 2.22 m/s

Problem 2 : A racehorse coming out of the gate accelerates from rest to a velocity of 14.0 m/s due in 1.90 s. What is its average acceleration?

Given v i = 0, v f = 14 m/s, Δt = 1.9 s

a = (v f – v i )/Δt = (14 m/s – 0)/1.9 s = 7.37 m/s

• What is Acceleration? – Khanacademy.org
• Acceleration – Physicsclassroom.com
• Acceleration – Physics.info
• Acceleration – Openstax.org

Article was last reviewed on Monday, January 2, 2023

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Acceleration

• A car is said to go "zero to sixty in six point six seconds". What is its acceleration in m/s 2 ?
• The driver can't release his foot from the gas pedal (a.k.a. the accelerator). How many additional seconds would it take for the driver to reach 80 mph assuming the aceleration remains constant?
• A car moving at 80 mph has a speed of 35.8 m/s. What acceleration would it have if it took 5.0 s to come to a complete stop?
• A baseball is pitched at 40 m/s (90 mph) in a Major League game. The batter hits the ball on a line drive straight toward the pitcher at 50 m/s (112 mph). Determine the the acceleration of the ball if it was in contact with the bat for 1 30  s.
• Which device(s) on a car can be used to control its acceleration?
• zero velocity, but non-zero acceleration
• zero acceleration, but non-zero velocity
• At main engine cutoff (MECO), the Space Shuttle is at an altitude of 113 km (70 miles), traveling 7,600 m/s (17,000 mph) relative to the Earth. This occurs 7 minutes 40 seconds into the mission. Determine the magnitude of the average acceleration experienced by the shuttle astronauts from lift off to MECO.
• The Flight of Fear is accelerated using linear induction motors (LIM), which generate a sequentially moving magnetic wave that propels the coaster like a surfer. A pair of LIMs is 85.3 m (280 foot) long and can accelerate the coaster to 24 m/s (54 mph) in 3.9 s. Determine the magnitude of the starting acceleration (in g) of the Flight of Fear.
• The HyperSonic XLC (Extreme Launch Coaster) was the world's first roller coaster to be launched using compressed air. Four, 150 kW (200 hp) compressed air motors accelerated the eight seat coaster from zero to 36 m/s (80 mph) in 1.8 s. Determine the magnitude of the starting acceleration (in g) of the Hypersonic XLC.
• When ejection seats were being developed, it was not known if a human could survive the intense acceleration needed to clear a jet fighter in an emergency. In 1954, US Air Force Colonel John Stapp was strapped into the seat of a rocket sled and blasted across the New Mexico desert at 282 m/s (632 mph) to examine the physiological effects of high speed ejection. The sled traveling at eight-tenths the speed of sound, a land speed record at that time, was then guided into a large trough of water, stopping it in a mere 1.4 s. Determine the magnitude of the average acceleration during the critical portion of this experiment. (Colonel Stapp subjected himself to several extreme acceleration experiments and survived all of them relatively unharmed.)
• US Federal Motor Vehicle Safety Standard No. 208 specifies performance requirements for the protection of vehicle occupants in crashes. In an auto accident, passengers and drivers should not experience an average head acceleration of more than 60 g for longer than 36 milliseconds. At what speed did the authors of this standard assume a typical, serious, survivable auto accident would take place?
• During a typical accident, a properly designed bicycle helmet should keep acceleration of the head below 200 g for a cumulative duration of three milliseconds and 150 g for a cumulative duration of six milliseconds. At what speed did the authors of this standard assume a typical accident would take place?
• A distressed car is rolling backward, downhill at 3.0 m/s when its driver finally manages to get the engine started. What velocity will the car have 6.0 s later if it can accelerate at 3.0 m/s 2 ?
• What zero-to-sixty time is equivalent to an average acceleration of 1 g?

How to Find Acceleration: Review and Examples

• The Albert Team
• Last Updated On: April 6, 2023

Acceleration is an important concept in physics used to describe motion and solve problems. In this article, we will define acceleration, the formula for acceleration, and its units. We will review examples of how to find acceleration including positive acceleration and negative acceleration.

What We Review

Review: Kinematic Terms

There are a few kinematic terms that you’ll need to know to understand acceleration. The first kinematic term is velocity. Velocity is the rate of change of position, or the displacement, over time. Displacement is another name for the change in position. Remember that an object moving at a constant velocity has a constant change of position every second. The motion map below shows an object moving with a constant velocity.

It is also important to remember the definition of vector quantities. Vectors are quantities that have both a magnitude, or size, and direction. For example, velocity is a vector quantity because it describes both how fast an object is moving (magnitude) and the direction the object is moving in. For an in-depth review of vectors, scalars, displacement, and velocity, visit our Albert blog post introducing kinematics. 

What is Acceleration?

Acceleration is defined as the rate of change of velocity. This means that if an object’s velocity is increasing or decreasing, then the object is accelerating. Unlike an object moving at a constant velocity, an accelerating object will not have a constant change in position every second. An accelerating object could cover more and more distance with every second, or less and less distance every second. The first motion map below shows an accelerating object that is speeding up and the second motion map shows an accelerating object that is slowing down.

Since acceleration depends on the change in velocity, acceleration is a vector quantity. This means that acceleration has both magnitude and direction. 

The standard units of acceleration are meters per second squared, \text{m/s}^2 . These units come from the units of velocity, meters per second, and the units of time, seconds. Since acceleration is the change in velocity over time, its units are the units of velocity (meters per second) divided by the units of time (second). 

For more information about acceleration and some examples, watch this quick video.

Acceleration Formula

Using our definition of what acceleration is, we can put together a formula for calculating acceleration.

Here, a is acceleration, \Delta v is the change in velocity, and t is time.

Before calculating acceleration, you will often need to first calculate the change in velocity. This is the difference between the object’s final velocity, v_f , and its initial velocity, v_i .

How to Find Acceleration Using the Acceleration Formula

In this next section, we will go over some examples of calculating acceleration. First, we will see an example of positive acceleration, then an example of negative acceleration. 

Example 1: Acceleration of a Car Speeding Up

A car starts from rest and moves forward while speeding up to 26\text{ m/s} in 8\text{ s} . Calculate the car’s acceleration. 

The first step in solving this problem is to determine the change in velocity. Since the car started from rest, its initial velocity, v_i , is 0\text{ m/s} . Its final velocity is 26\text{ m/s} . Therefore, the car’s change in velocity is:

Now we can calculate the car’s acceleration by dividing this change in velocity by the time of 8\text{ s} :

Example 2: Acceleration of a Car Slowing Down

Now let’s consider a situation where an object is slowing down. 

A car initially moving forward at a velocity of 26\text{ m/s} approaches a school and slows down to a velocity of 11\text{ m/s} in 3\text{ s} . Calculate the car’s acceleration.

The first step is to find the change in velocity. The car’s initial velocity is 26\text{ m/s} and its final velocity is 11\text{ m/s} . Therefore, the car’s change in velocity is:

Now we can calculate the car’s acceleration by dividing this change in velocity by the time of 3\text{ s} :

We will explain in the next section more about what a negative acceleration means. 

Determining the Direction of Acceleration

As a vector quantity, acceleration has both magnitude and direction. The direction of acceleration depends on if the object is speeding up or slowing down, and the direction the object is moving. In general, if an object is speeding up, its acceleration will be in the same direction as its motion. If an object is slowing down, its acceleration is in the opposite direction of its motion. 

Positive Acceleration

There are two types of situations where an object can have a positive acceleration. If an object is speeding up and moving in a positive direction, it has a positive acceleration. The car speeding up in the first example was an example of positive acceleration. The car is moving forward in a positive direction and speeding up, so the acceleration is in the same direction as the car’s motion. 

An object can also have a positive acceleration if it is slowing down while moving in a negative direction. Since the object is slowing down, the acceleration is in the opposite direction of its motion. 

Negative Acceleration

As we saw in the second example, an object can have a negative acceleration when the object is slowing down while moving in a positive direction. The car in the school zone was moving forward in a positive direction and slowing down, so the acceleration was in the opposite direction of the car’s motion. 

An object can also have a negative acceleration if it is speeding up while moving in a negative direction. Since the object is speeding up, its acceleration is in the same direction as its motion. 

Examples: Identify the Direction of Acceleration

Example 1: a cyclist.

Consider a cyclist riding on a straight road. The cyclist moves in a positive direction at a constant velocity. Suddenly, the cyclist sees an obstacle and applies the brakes, slowing down until they come to a complete stop.

To find the direction of acceleration for the cyclist, first, identify whether the cyclist is speeding up or slowing down. In this case, the cyclist is slowing down. Since the cyclist is slowing down, the acceleration will be in the opposite direction of the motion. The cyclist is moving in a positive direction, so the cyclist’s acceleration must be in a negative direction.

Example 2: A Soccer Player

A soccer player is running backward (in the negative direction) to receive a pass from a teammate. Once they receive the ball, they quickly change direction and start running forward, speeding up as they go.

To determine the direction of acceleration for the soccer player, consider two parts of their motion:

• When the player is running backward in a negative direction: Since the player is slowing down and moving in a negative direction, the acceleration will be in the opposite direction of the motion (positive direction).
• When the player starts running forward in a positive direction: In this case, the player is speeding up while moving in a positive direction, so the acceleration will be in the same direction as the motion (positive direction).

In this scenario, the soccer player experiences positive acceleration in both stages, even though they are initially moving in a negative direction.

Review more examples with the Physics Classroom’s Acceleration Concept Builder.

Practice Using the Acceleration Formula in Word Problems

Now that we have reviewed acceleration and how to find both its magnitude and direction, we can apply this knowledge to solve more complicated physics word problems. 

Example 1: Acceleration of a Falling Ball

A ball dropped from rest reaches a downward velocity of 24.5\text{ m/s} after falling for 2.5\text{ s} . What is the magnitude and direction of the ball’s acceleration?

To find the magnitude of the acceleration, we will substitute the given values into the formula for acceleration. Since the ball starts from rest, its initial velocity, v_i is 0\text{ m/s} . 

Since the ball is speeding up, the acceleration will be in the same direction as the ball’s motion. The ball is moving downward in a negative direction. Therefore the direction of the acceleration is also downward or negative. In this case, this is the free fall acceleration and can be expressed as either -9.8\text{ m/s}^2 or 9.8\text{ m/s}^2 downward. 

Example 2: Final Velocity of a Skateboarder

A skateboarder with an initial speed of 2\text{ m/s} accelerates down a ramp at a rate of 6\text{ m/s}^2 for 1.5\text{ s} . What is the skateboarder’s final velocity? 

The first step to solve this problem is to use the formula for acceleration to determine the skateboarder’s change in velocity. Let’s substitute the values for acceleration and time into the formula:

Now, we can multiply both sides by the time to solve for the change in velocity:

Change in velocity is the difference between the object’s final and initial velocities. Using our change in velocity and the skateboarder’s initial velocity produces:

The last step to solve for the final velocity is to add the initial velocity to the change in velocity: 

Therefore, the skateboarder’s final velocity is 11\text{ m/s} directed down the ramp. 

Kinematic Equations

Now that you know all four kinematic terms (time, displacement, velocity, and acceleration), you will be able to describe the motion of objects. For objects with uniform acceleration, the relationships between these variables are expressed through kinematic equations.

The first kinematic equation is actually just a variation of the formula for acceleration solved for the final velocity. These equations will allow you to predict the motion of objects and solve for unknown variables in physics problems. 

Acceleration is the rate of change of velocity and allows us to describe the motion of objects with changing velocities. It is important to remember that acceleration is a vector quantity with both magnitude and direction. To identify if an object’s acceleration is positive or negative, we need to consider both if the object is speeding up or slowing down and the direction the object is moving. 

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High School Physics

Average Acceleration Formula 

Last updated on June 2nd, 2023 at 09:09 am

In this post, we have taken the Average Acceleration Formula as our topic of discussion. We will see how to define and find it, and how to solve sample numerical problems using the formula. This post is useful for class 11 students who are studying this chapter.

When velocity changes an object is said to undergo acceleration. Quantitatively, we define acceleration as the rate of change of velocity, just as we defined velocity as the rate of change of position.

Definition of Average Acceleration

The average acceleration over a time interval Δ t is the ratio of change in velocity Δ v (that takes place over the time interval Δ t ) and the time interval Δ t itself.

Average Acceleration Formula

Average Acceleration a avg = Δ v / Δ t where Δ v is the change in velocity over the time interval Δ t .

Also Read about Instantaneous Acceleration – formula, definition, numerical problems

How to find average acceleration with the formula

– For a time duration Δt, we have to get the initial and final velocity. – Then we have to find out the difference between the final velocity and the initial velocity. Use this one to do that, Δv = V final  –  V initial – Now, we have to calculate the ratio of Δv and Δt. [ as we know, a avg = Δv / Δt ] – This will give us the average acceleration value. If you find the value negative that means it’s actually retardation.

Sample numerical problems on average acceleration – all solved

1 ) A car is moving with a velocity of 20 m/s. The driver accelerated it for 10 seconds and reached a velocity of 40 m/s. What is the average acceleration?

Solution: Initial velocity = 20 m/s and final velocity is 40 m/s. And the elapsed time = 10 seconds Therefore, a = Δv / Δt = (40 -20) / 10 m/s^2 = 2 m/s^2

2 ) A car is moving with a velocity of 30 m/s. The driver applied brake for 5 seconds to bring it down to zero. What is the average acceleration? Is it retardation?

Solution: Initial velocity =30 m/s and final velocity is 0 m/s. And the elapsed time = 5 seconds Therefore, a = Δv / Δt = (0-30) / 5 m/s^2 = -6 m/s^2 The negative value of a tells that it’s actually retardation.

Also Read: (suggested reading)

Difference between Instantaneous Speed and Instantaneous Velocity Difference between average speed and average velocity Instantaneous Velocity – definition & equation with solved problem Average velocity – definition, formula Instantaneous Acceleration – definition & formula with solved problem Average Acceleration and its formula & solved numerical problems

Related Posts:

• Average Acceleration Calculator
• Average velocity - definition, formula, numerical problem
• Difference between average speed and average velocity
• Instantaneous Acceleration - definition & formula with solved problem
• How uniform acceleration is different from nonuniform acceleration?
• Average Velocity Calculator - using Displacement

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Acceleration Formula

What is Acceleration?

When a stationary car starts suddenly, we get pushed up backward, and when brakes are applied, we get pushed forward against our seat, or when our car takes a sharp right turn, we get pushed towards the left. We experience these situations because our car is accelerating.

Simply when there is a change in Velocity, there will be Acceleration. Let’s understand the concept of Acceleration with illustrative examples.

Let’s suppose I have a car moving with a constant Velocity of 90 kmph along a straight line. I can see a helicopter flying at roughly a speed of 20,000 kmph. If I were to ask you that in these two cases, where do you find the Acceleration? Your answer will be surely no because both are moving at a constant pace, so no Acceleration in both cases. 

Now, if I ask you that Acceleration is equal to high speed. What will be your answer? You may say yes, but that’s not true for sure.  Want to know why?  It’s because Acceleration is the rate of change of Velocity. Now, let’s understand the Acceleration formula.

General Formula of Acceleration

We already know that Velocity is a speed with direction; therefore, it is a vector quantity. The Acceleration ‘a’ is given as:

      \[ a  = \frac{\text{Change in Velocity}}{\text{Time Taken}}\]

This formula states that the rate of change in Velocity is the Acceleration, or if the Velocity of an object changes from its initial value ‘u’ to the final value ‘v’, then the expression can be simply written as:

                         \[a =  \frac{(v - u)}{t}\]        

Acceleration Formula in Physics 

In Physics , Acceleration is described as the rate of change of Velocity of an object, irrespective of whether it speeds up or slows down. If it speeds up, Acceleration is taken as positive and if it slows down, the Acceleration is negative. It is caused by the net unbalanced force acting on the object, as per Newton’s Second Law. Acceleration is a vector quantity as it describes the time rate of change of Velocity, which is a vector quantity. Acceleration is denoted by a. Its SI unit is \[\frac{m}{s^{2}}\] and dimensions are \[[M^{0}L^{1}T^{–2}]\].

If \[v_{0}, v_{t}\] and t represents the initial Velocity, final Velocity and the time taken for the change in Velocity, then, the Acceleration is given by:

\[\overrightarrow{a} = \frac{\overrightarrow{v_t} - \overrightarrow{v_0}}{t}\]

In one dimensional motion, we can use;

\[a = \frac{v_t - v_0}{t}\]

If \[\overrightarrow{r} \]represents displacement vector and \[\overrightarrow{v} = \frac{\overrightarrow{\text{d}r}}{\text{d}t}\] represents the velocity, then;

Acceleration:  \[\overrightarrow{a} = \frac{\overrightarrow{\text{d}v}}{\text{d}t} = \frac{\overrightarrow{\text{d}^{2}v}}{\text{d}^{2}t}\]

In one dimensional motion, where x is the displacement, and \[v = \frac{\text{d}r}{\text{d}t}\] is the Velocity, then;

\[a = \frac{\text{d}{v}}{\text{d}t} = \frac{\text{d}^{2}x}{\text{d}^{2}t}\]

A car starts from rest and achieves a speed of 54 \[\frac{km}{h}\] in 3 seconds. Find its Acceleration?

Solution:  \[v_0\] = 0, \[v_t\] = 54 \[\frac{km}{h}\] = 15 \[\frac{m}{s}\], t = 3s, a = ?

Acceleration:  

\[a = \frac{v_t - v_0}{t} = \frac{15 - 0}{3} = 5 \frac{m}{s^{2}}\] 

A body moves along the x- axis according to the relation \[x = 1 – 2 t + 3t^{2}\], where x is in meters and t is in seconds. Find the Acceleration of the body when t = 3 s

We have: \[x = 1 – 2 t + 3t^{2}\]

then; Velocity \[v = \frac{\text{d}x}{\text{d}t} = -2 + 6t \]

Acceleration:  \[v = \frac{\text{d}v}{\text{d}t} =  6  \frac{m}{s^{2}}\].

(We see that the Acceleration is constant here. Therefore, at t = 3s also, its value is 6  \[\frac{m}{s^{2}}\]).

Solved Questions Using Acceleration Formula:

1. What will be the Acceleration of a Car if it Slows from 90 \[\frac{km}{h}\] to a Stop in 10 sec? 

Here, u = 90 \[\frac{km}{h}\] = \[ \frac{90 \times 5}{18} = 25 \frac{m}{s^{2}} \] because initially it was moving at a speed of 90 kmph then reached zero.

Final Velocity ‘v’ = 0 kmph, and t  = 10 seconds

Now, applying the formula here:

\[a  = \frac{(0 - 25)}{10}  = (-) 2.5 \frac{m}{s^{2}} \]

2. A Girl Starts her Motion in a Straight Line at a Velocity of 30 \[\frac{m}{s}\], her Velocity is Changing at a Constant Rate. If She Stops after 60 s, What is her Acceleration?

Answer: Here, the initial Velocity of a girl was 30 \[\frac{m}{s}\] and stops, so her final Velocity will become 0 m/s. Now, the deceleration or retardation occurs, which is just the opposite of Acceleration and it can be determined as:

                  \[  a  = \frac{(0 - 30)}{60} = (-) 0.5 \frac{m}{s^{2}} \]

Question 3: A Car Moves in a Circular Track with a Constant Velocity; will it Experience Acceleration?

Answer: Here, the speed is constant; however, the direction is continuously varying, which means the Velocity is also varying. It states that the car will experience Acceleration.

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FAQs on Acceleration Formula

1. Where can I find sums on Acceleration?

You can find ample examples and sums on Acceleration if you choose to read Acceleration Formula with examples and solved problems on Vedantu. This has a detailed description of what Acceleration is, what its examples are, how it's related to speed and direction.  You will find the kind of sums that will come for the tests here so that you can practice them and secure good marks. Do not skip reading any part as that might cause problems with understanding the concepts later on.

2. How do I clear my concepts related to Velocity and speed?

You should understand the very basics of what each is so that you can answer questions when they are asked. Go through Acceleration Formula with examples and solved problems available on this page on Vedantu as this is the ideal guide for those who are looking to clear their concepts. 

Velocity is the speed with direction. The rate of change in Velocity would be termed Acceleration.  Understanding this is key to solving the sums that come from this chapter and topic. You should go through the entire description along with its formulae and examples. Practice a few sums and then assess your understanding.

3. When a car suddenly halts, why do we get pushed forward against our seat?

As a car suddenly stops,  we tend to get pushed forward as our car starts accelerating. As there is a change in the speed with direction (Velocity), Acceleration occurs.  More about this is provided if you go to  Acceleration Formula with examples and solved problem. This can be found on Vedantu and contains all the relevant details that need to be known. It has explained everything in an extremely simplified manner with the help of examples and is an ideal guidebook for those who are looking to clear their concepts related to Acceleration.

4. Is Acceleration the same as very high speed?

You may be partially correct in stating so but not entirely. Acceleration is more essentially the rate of change of Velocity.  You can read from Acceleration Formula with examples and solve problems on Vedantu and then understand better. It has the description of Acceleration along with its sums and a lot of examples. It has a comprehensive description of the topic and its related concepts so that students do not get confused while learning about the topic. Acceleration as a concept is crucial for the higher classes as the Science and Maths study material that they come across then, will be based on these.

5. Can I skip the chapter and the sums on Acceleration when I am studying?

You should not skip anything at any point in time as it could lead to securing lower grades overall. If you have any queries regarding the concepts on Acceleration and its related problems, you can refer to Acceleration Formula with examples and solved problems on Vedantu’s platform. This will clear all your doubts so that you do not feel tempted to skip this chapter just because it's slightly tricky. Once you have completely scanned this page, you will be able to solve the sums that come for your tests and be more confident in tackling questions that come from the chapter.

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Average Acceleration: Solved Examples with Definition

In this article, some average acceleration worked examples are provided for high school physics.

Definition of Average Acceleration:

Objects move either with constant or varying velocities. For example, in everyday experience, you noticed when you step on the gas pedal of your car or brake one, the car's speed increases or decreases, Or in a turn your car's direction changes.

These changes in the magnitude of velocity or direction of a moving body, in physics, are described by acceleration.

In other words, the average acceleration is defined as the rate of change of the object's velocity and is given by the following formula: \[\text{Average Acceleration}\equiv\frac{\text{change in velocity}}{\text{time elapsed}}\] Typically average acceleration is denoted by $\bar a$ or $\vec{a}_{\text{ave}}$.

The units for acceleration are meters per squared second ($\rm \dfrac{m}{s^2}$).

Average Acceleration Examples:

Example (1): An automobile slows its speed from $25\,\rm m/s$ to $15\,\rm m/s$ in a time interval $2.5\,\rm s$, what is its acceleration?

Solution : the change in velocity of the automobile, $\Delta v=15-25=-10\,\rm m/s$, occurs in a time period of $\Delta t=2.5\,\rm s$. Thus, its average acceleration is found to be \[\bar{a}=\frac{-10}{2.5}=\boxed{-4\,\rm m/s^2}\] If we take the positive direction to be the right, then the average acceleration points to the left in the opposite direction to the velocity, meaning that the automobile has a slowing down motion. This is also an example of a decelerating motion. 

One of the features of this motion is the moving object eventually comes to a rest or reverses its direction.

Keep in mind that the average acceleration is always in the same direction as the change in the velocity, but not always in the direction of the motion.  

As you can see in this average acceleration example, the automobile is traveling to the right (the direction of the motion), but the direction of its average acceleration and change in velocity is toward the left. 

Example (2): If a runner starts at rest and reaches a speed of $20\,\rm m/s$ in $2\,\rm s$, what is his average acceleration? 

Solution : start at rest, so $v_1=0$, and finally reach a speed of $v_2=20\,\rm m/s$. Thus, the average acceleration is \[\bar{a}=\frac{20-0}{2}=\boxed{10\,\rm m/s^2}\] Assume the positive direction to be the right and the runner is moving in this direction. Its average acceleration is obtained positively, meaning the average acceleration is in the same direction as the motion.

Example (3): A car is accelerating from $20\,\rm m/s$ to rest in $1.0\,\rm s$. What is its average acceleration? 

Solution : The initial and final velocities of the car, $v_1=20\,\rm m/s$ and $v_2=0$, in a time interval $1\,\rm s$ are given. Thus, the change in the car's velocity is $\Delta v=0-20=-20\,\rm m/s$.

According to the average acceleration formula, we have \[\bar{a}=\frac{-20}{1}=\boxed{-20\,\rm m/s^2}\] Assuming the positive direction to be the right and the car is moving toward this direction, the negative of average acceleration means that it points to the opposite direction of the motion. Thus, the car eventually will come to a stop, as expected. 

Example (4): A car is traveling in a straight line along a highway at a constant speed of $80$ miles per hour for 10 seconds. Find its acceleration?

Solution: Average acceleration is a change in velocity divided by the time taken. Since the car's velocity (magnitude and direction) is constant over the entire course, so by definition of average acceleration, it is zero, i.e., $\bar{a}=0$. 

Example (5): A plane has a take-off speed of $300\,{\rm \frac{km}{h}}$. What is the average acceleration (in $\rm \frac{m}{s^2}$) of the plane if the plane started from rest and took 45 seconds to take off? 

Solution: Plane is initially at rest so $\vec{v_1}=0$ and its take off speed is $\vec{v_2}=300\,{\rm \frac{km}{h}}$. First off, convert ${\rm \frac{km}{h}}$ to SI units of velocity ${\rm \frac{m}{s}}$ as below\begin{align*} \rm{300\,\frac{km}{h}}&=\rm{300\,\frac{1000\,m}{3600\,s}}\\\\&=\rm{300\,\frac{1000}{3600}\,\frac{m}{s}}\\\\ &= \rm{83.4\,\frac{m}{s}} \end{align*} Now ratio of change in velocity, $\Delta \vec{v}=83.4\,{\rm \frac ms}$ over time elapsed $\Delta t=45\,\rm s$ is definition of average acceleration.\begin{align*} \bar{a}&=\frac{\Delta \vec{v}}{\Delta t} \\\\ &=\frac{83.4\,\rm{m/s}}{40\,{\rm s}}=2.085\,\rm{\frac{m}{s^2}} \end{align*} 

Example (6): What average acceleration is needed to accelerate a car from $36\,{\rm \frac{km}{h}}$ to $72\,{\rm \frac{km}{h}}$ in $25$ seconds?

Solution: Initial and final velocities are $36\,{\rm \frac{km}{h}}$ and $72\,{\rm \frac{km}{h}}$, respectively. As before, convert them in SI units of the velocity as follows \begin{align*} \rm{\frac{km}{h}} &= \rm{\frac{1000\,m}{3600\,s}}\\ \\&= \rm{\frac{1000}{3600}\,\frac{m}{s}}\\\\ &= \rm{\frac{10}{36}\,\frac{m}{s}} \end{align*} In other words, multiply them by $\frac{10}{36}$. Then, $v_1=36\times\frac{10}{36}=10\,{\rm \frac ms}$ and $v_2=72\times\frac{10}{36}=20\,{\rm \frac ms}$. Now, by dividing the change of velocity, $\Delta v=20-10=10\,{\rm \frac ms}$ over the time elapsed $\Delta t=25\,{\rm s}$, we get the desired average acceleration \begin{align*} \bar{a}&=\frac{\Delta \vec{v}}{\Delta t} \\\\ &=\frac{20\,\rm{m/s}}{25\,{\rm s}}\\\\&=\boxed{0.8\,\rm m/s^2} \end{align*}

Example (7): Starting with a constant velocity of $50\,{\rm \frac{km}{h}}$, a car accelerates for 32 seconds at an acceleration of $0.5\,{\rm \frac{m}{s^2}}$. What is the velocity of the car at the end of the period of $32$ seconds of acceleration?

Solution:  Here, the initial velocity, acceleration, and time interval over which this car accelerates are given, and the final velocity is requested. Therefore, using the definition of average acceleration we get, \begin{align*} \bar{a} &=\frac{\vec{v_2}-\vec{v_1}}{\Delta t} \\\\ 0.5 &= \frac{v_{2}-50\times\frac{10}{36}}{32}\\\\ 32\times 0.5 &= v_{2}-50\times\frac{10}{36}\\\\ \Rightarrow v_2 &=16+13.9=29.9\,\rm m/s \end{align*} For the second equality refer to Note $4$. 

Example (8): A ball weighing $50\,\rm g$ with a speed of $25\,\rm m/s$ strike a wall and rebounds at $22\,\rm m/s$. Assuming the ball is in contact with the wall for $3.5\,\rm ms$, what is the magnitude of the average acceleration of the ball during this time interval?

Solution : Average acceleration is defined as the change in velocity divided by the time interval that this change occurs. Substituting the given numerical values into the average velocity formula get \begin{align*} \bar{a}&=\frac{v_2-v_1}{t_2-t_1}\\\\&=\frac{22-25}{3.5\times 10^{-3}} \\\\ &=\boxed{857.14\,\rm m/s^2} \end{align*} This is a very large acceleration. 

Example (9): How long does it take to accelerate a car from a speed of $50\,{\rm \frac{km}{h}}$ to a speed of $100\,{\rm \frac{km}{h}}$ at a rate of $1\,{\rm \frac{m}{s^2}}$?

Solution:  In this example, the unknown quantity is the time interval over which average acceleration occurs. Thus from definition of average acceleration along a straight line, we have \begin{align*} \bar{a} &=\frac{\vec{v_2}-\vec{v_1}}{\Delta t} \\\\ 1&=\frac{\left(100-50\right)\times \frac{10}{36}}{t}\\\\ \Rightarrow t&=\frac{\left(100-50\right)\times \frac{10}{36}}{1}\\\\ &=13.9\,{\rm s} \end{align*} In the second equality, kilometer/hour ($\rm km/h$) is converted into the meter/second ($\rm m/s$) by multiplying by $\frac{10}{36}$.

It is also possible to compute the acceleration of a moving object using a position versus time graph .

Example (10): A car start moving with a velocity of $-4\,{\rm m/s}$ from position $+4\,{\rm m}$. After $2\,{\rm s}$, its position is $-1\,{\rm m}$ with final velocity $-1\,{\rm m/s}$. Find: (a) What is the displacement of the car? (b) What is the average velocity of the car? (c) What is the average acceleration of the car?

Solution:  (a) Displacement is defined as the change in position of an object \begin{align*}\Delta x&=x_2-x_1\\&=-1-4=-5\,{\rm m}\end{align*} (b) Average velocity is $\bar{v}=\frac{\Delta x}{\Delta t}$ so \begin{align*}\bar{v}&=\frac{\Delta x}{\Delta t}\\\\ &=\frac{-5}{2}\\\\ &=-2.5\,{\rm m/s}\end{align*} (c) By applying average acceleration formula $\bar{a}=\frac{\Delta v}{\Delta t}$, we get \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\ &=\frac{-1-(-4)}{2} \\\\ &=+1.5\,{\rm m/s^2}\end{align*}In this example, the negatives of velocities indicates the car moves in the negative $x$ direction but the average acceleration is toward the positive $x$ direction. Consequently, the car is slowing down because $a.v<0$, so the motion is slowing down.

All these kinematics quantities are also found using a position vs. time graph .

Example (11): A $5-g$ ball is dropped from a height. It bounces off a brick wall at $10\,{\rm m/s}$ and rebounds at $8\,{\rm m/s}$. The ball is in contact with the wall for $2\,{\rm ms}$. Find the average acceleration of the ball during the contact time.

Solution: Let the positive direction be up so the given data are $v_1=-10\,{\rm m/s}$, $v_2=+8\,{\rm m/s}$ and $\Delta t=2\times 10^{-3}\,{\rm s}$ which we converted the milliseconds into seconds. 

By applying average acceleration formula, we get \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{v_2-v_1}{\Delta t}\\\\&=\frac{+8-(-10)}{2\times 10^{-3}}\\\\&=9\times 10^{3}\,{\rm m/s^2}\end{align*}

Example (12):  A plastic bullet of $2\,{\rm g}$ from a rifle is fired at a brick wall. It strikes the wall with $20\,{\rm m/s}$ and returns at $18\,{\rm m/s}$. If the contact time of the ball with the wall is $3\,{\rm ms}$, what is the bullet's average acceleration?

Solution: suppose the positive direction is toward the positive $x$ direction. Thus, we have $v_1=+20\,{\rm m/s}$, $v_2=-18\,{\rm m/s}$ and $\Delta t=3\times 10^{-3}\,{\rm s}$. Therefore, the average acceleration definition gives \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{v_2-v_1}{\Delta t}\\\\&=\frac{-18-(+20)}{3\times 10^{-3}}\\\\&=-12\times 10^{3}\,{\rm m/s^2}\end{align*} The minus sign of acceleration indicates its direction which is toward the negative $x$ axis.

Example (13): The velocity-time graph of a moving car along a straight path is shown below. Find its average acceleration during the first 2 seconds of motion.

Solution : the slope of a line segment connecting two points on a velocity-time graph gives us the average acceleration. 

The slope is also defined as the change in the vertical axis divided by a change in the horizontal axis. The starting point has coordinate $(v=-4\,{\rm m/s},t=0)$ and final point is $(v=0,t=2\,{\rm s})$. So, the slope of line joining these points is found as below \begin{align*}\text{slope=acceleration}&=\frac{\Delta x}{\Delta t}\\\\&=\frac{0-(-4)}{2-0}\\\\&=+2\quad {\rm m/s^2}\end{align*} Thus, the car increases its speed at a constant rate of $2\,{\rm m/s^2}$ in 2 s. 

Average Acceleration Examples Tips:

Note $1$: Velocity is a vector quantity, meaning it has both a magnitude and a direction. If one of those changes over a period of time, then we have acceleration.  Recall from uniform circular motion problems that an object with a uniform (constant) speed in a circular path can still have an acceleration like satellites around the Earth that move constantly but undergoes a type of acceleration called centripetal acceleration.  (On the page of  vector practice problems , you can find how to calculate the magnitude and direction of a given vector)

Note $2$: Change in velocity is defined as the difference between initial ($\vec{v_1}$) and final ($\vec{v_2}$) velocities as below \[\Delta\vec{v}=\vec{v_2}-\vec{v_1}\]

Note $3$:  If an object moves in a straight line since in this case there is no change in direction, the only part of the velocity vector changes is its magnitude. Thus, we can omit its direction sign, i.e., $\vec v=v$. 

Note $4$: The average acceleration vector points in the same direction as the vector $\Delta \vec{v}$.

According to vector definition in physics , velocity is a vector quantity with a magnitude and direction. If one of those changes during a finite period of time $\Delta t$, then we have average acceleration (vector) and is given by the following formula: \begin{align*} \vec{a}_\text{ave}&=\frac{\text{change in velocity vector}}{\text{elapsed time}}\\ &=\frac{\Delta \vec{v}}{\Delta t}\\ &=\frac{\vec{v}_2-\vec{v}_1}{t_2-t_1} \end{align*} SI unit of acceleration is meter per second per second ($m/s^2$) and its dimension is ${\rm L\,T^{-2}}$. 

Practicing more problems on speed, velocity, and acceleration  also helped a better understanding of the average acceleration concept.

Author: Dr. Ali Nemati Last Update: Oct 1, 2022

© 2015 All rights reserved. by Physexams.com

OpenAI Scale Ranks Progress Toward `Human-Level’ Problem Solving

By Rachel Metz

OpenAI has come up with a set of five levels to track its progress toward building artificial intelligence software capable of outperforming humans, the startup’s latest effort to help people better understand its thinking about safety and the future of AI.

The ChatGPT maker, seen by many as a leader in the race to build more powerful AI systems, shared the new classification system with employees on Tuesday during an all-hands meeting, an OpenAI spokesperson said. The tiers, which OpenAI plans to share with investors and others outside the company, range from the kind of AI available today that can interact in conversational language with people (Level 1) to AI ...

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Welcome to the daily solving of our PROBLEM OF THE DAY with Saurabh Bansal. We will discuss the entire problem step-by-step and work towards developing an optimized solution. This will not only help you brush up on your concepts of Arrays but also build up problem-solving skills. Given an array, arr of integers, and another number target , find three integers in the array such that their sum is closest to the target. Return the sum of the three integers.

Note: If there are multiple solutions, return the maximum one.

Input: arr[] = [-7, 9, 8, 3, 1, 1], target = 2 Output: 2 Explanation: There is only one triplet present in the array where elements are -7,8,1 whose sum is 2.

Give the problem a try before going through the video. All the best!!! Problem Link: https://practice.geeksforgeeks.org/problems/three-sum-closest/1

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