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# Constant Acceleration

Acceleration is defined as the change in velocity over time. If the rate of change of velocity of a body remains constant over time, it is known as constant acceleration.

A ball dropped from a height falling freely under the force of gravity with no other external force acting on it will be falling with constant acceleration equal to the acceleration due to gravity.

In reality, it is very difficult to realise perfect constant acceleration. This is because there will always be multiple forces acting on an object. In the above example, various atmospheric forces such as air resistance will also be acting on the ball. However, the variations in the resultant acceleration might be small enough that we can still model its motion using the concepts of constant acceleration.

## Constant acceleration graphs

It is possible to graphically represent the motion of an object. In this section, we will look at two types of graphs that are commonly used for representing the motion of an object moving with constant acceleration:

1. Displacement-time graphs

2. Velocity-time graphs

### Displacement-time graphs

The motion of an object can be represented using a displacement-time graph.

Displacement is represented on the Y-axis and time (t) on the X-axis. This implies that the change of position of the object is plotted against the time it takes to reach that position.

Here are a few things to keep in mind for displacement-time graphs:

• Since velocity is the rate of change of displacement, the gradient at any point gives the instantaneous velocity at that point.

• Average velocity = (total displacement)/(time taken)

• If the displacement-time graph is a straight line, then the velocity is constant and the acceleration is 0.

The following displacement-time graph represents a body with a constant velocity, where s represents the displacement and t the time taken for this displacement.

Displacement-time graph for a body moving with a constant velocity, Nilabhro Datta, Study Smarter Originals

The following displacement-time graph represents a stationary object with zero velocity.

Displacement-time graph for a body having zero velocity, Nilabhro Datta, Study Smarter Originals

The following displacement-time graph represents an object moving with constant acceleration.

Displacement-time graph for a body moving with a constant acceleration, Nilabhro Datta, Study Smarter Originals

### Velocity-time graphs

The motion of an object can also be represented using a velocity-time graph. Customarily, the velocity (v) is represented on the Y-axis and time (t) on the X-axis.

Here are a few things to keep in mind for velocity-time graphs:

• Since acceleration is the rate of change of velocity, in a velocity-time graph the gradient at a point gives the acceleration of the object at that point.

• If the velocity-time graph is a straight line, then the acceleration is constant.

• The area enclosed by the velocity-time graph and the time-axis (horizontal axis) represents the distance travelled by the object.

• If the motion is in a straight line with positive velocity, then the area enclosed by the velocity-time graph and the time-axis also represents the displacement of the object.

The following velocity-time graph represents the motion of a body moving with a constant velocity and therefore zero acceleration.

Velocity-time graph for a body moving with constant velocity, Nilabhro Datta, Study Smarter Originals

As we can see, the value of the velocity component remains constant and does not change with time.

The following graph depicts the motion of a body moving with constant (non-zero) acceleration.

Velocity-time graph for a body moving with constant acceleration, Nilabhro Datta, Study Smart Originals

We can see how in the above graph, the velocity is increasing at a constant rate. The slope of the line gives us the acceleration of the object.

## Constant acceleration equations

For a body moving in a single direction with constant acceleration, there is a set of five commonly used equations that are used to solve for five different variables. The variables are:

1) s = displacement

2) u = initial velocity

3) v = final velocity

4) a = acceleration

5) t = time taken

The equations are known as the constant acceleration equations or the SUVAT equations.

### The SUVAT Equations

There are five different SUVAT equations that are used to connect and solve for the variables above in a system of constant acceleration in a straight line.

1) v = u + at

2) s = ½ (u + v) t

3) s = ut + ½at²

4) s = vt - ½at²

5) v² = u² + 2 as

Note that each equation has four of the five SUVAT variables. Thus given any of the three variables, it would be possible to solve for any of the other two variables.

A car starts accelerating at 4 m / s² and crashes into a wall at 40 m / s after 5 seconds. How far was the wall when the car started accelerating?

#### Solution

Here v = 40 m / s, t = 5 seconds, a = 4 m / s².

s = vt - ½at²

Solving for s you get:

s = 40 × 5 - ½ × 4 × 5² = 150 m

A driver applies the brakes and his car goes from 15 m / s to a halt within 5 seconds. How much distance did it travel before coming to a halt?

#### Solution

Here u = 15 m / s, v = 0 m / s, t = 5 seconds.

s = ½ (u + v) t

Solving for s:

s = ½ (15 + 0) 5 = 37.5 m

## Constant acceleration due to gravity

The force of gravity exerted by the Earth causes all objects to accelerate towards it. As we have already discussed, an object falling from a height falls with practically constant acceleration. If we ignore the effects of air resistance and the almost negligible gravitational pull of other objects, this would be perfectly constant acceleration. The acceleration due to gravity also does not depend on the mass of the object.

The constant g is used to represent the acceleration due to gravity. It is approximately equal to 9.8 m / s². If you are solving problems that require you to use the value of acceleration due to gravity, you should use the value g = 9.8 m / s² unless a more accurate measurement is provided to you.

A body falling from a height can be considered a body accelerating at a rate of g. A body being thrown up with an initial velocity can be considered a body decelerating at a rate of g until it reaches its peak height where the acceleration is zero. When the object falls after reaching its peak height, it will accelerate again at a rate of g while going down.

A cat sitting on a wall that is 2.45 meters high sees a mouse on the floor and jumps down trying to catch it. How long will it take for the cat to land on the floor?

#### Solution

Here u = 0 m / s, s = 2.45m, a = 9.8 m / s².

s = ut + ½at²

Substituting all values to solve for t:

=> 2.45 = 0 × t + ½ × 9.8 × t²

=> 2.45 = 4.9t²

=> t = 1 / √2 = 0.71 seconds

A ball is thrown up with an initial velocity of 26 m / s. How long will it take the ball to reach its peak height? Assume g = 10 m / s².

#### Solution

Here u = 26 m / s, v = 0 m / s, a = -10 m / s².

v = u + at

Substituting all values in the equation:

=> 0 = 26 - 10t

Solving for t

=> t = 2.6 seconds

## Constant Acceleration - Key takeaways

• Acceleration is the change in velocity over time. If the rate of change of velocity of a body remains constant over time, it is known as constant acceleration.

• The motion of an object can be represented graphically. Two commonly used types of graphs for this purpose are displacement-time graphs and velocity-time graphs.

• There are five common equations of motion used in a system involving constant acceleration in a straight line. These are commonly known as the SUVAT equations.

• A body falling from a height can be considered a body accelerating at a rate of g (constant of acceleration due to gravity). A body being thrown up with an initial velocity can be considered a body decelerating at a rate of g until it reaches its peak height.

The acceleration due to gravity is constant for all objects close to the Earth’s surface as it depends on the mass of the Earth which is a constant.

Acceleration is the change in velocity over time. If the rate of change of velocity of a body remains constant over time, it is known as constant acceleration.

You can calculate constant acceleration by dividing the change in velocity by the time taken. Therefore, a = (v – u)/t, where a = acceleration, v = final velocity, u = initial velocity and t = time taken.

Velocity is the displacement per unit time, whereas acceleration is the change in that velocity per unit time.

There are five commonly used equations for motion with constant acceleration

1)    v = u + at

2)    s = ½ (u + v) t

3)    s = ut + ½at²

4)    s = vt - ½at²

5)    v² = u² + 2 as

where s= Displacement, u= Initial velocity, v= Final velocity, a= Acceleration, t= Time taken.

## Final Constant Acceleration Quiz

Question

What are the five SUVAT variables ?

s= Displacement, u= Initial velocity, v= Final velocity, a= Acceleration, t= Time taken

Show question

Question

You are provided s= Displacement, u= Initial velocity, v= Final velocity, and asked to find acceleration. Which equation should you use?

v² = u² + 2 as

Show question

Question

You are provided s= Displacement, u= Initial velocity, a= Acceleration, and asked to find the time taken. Which equation should you use?

s = ut + ½at²

Show question

Question

You are provided u= Initial velocity, v= Final velocity, t= Time taken and asked to find displacement. Which equation should you use?

s = ½ (u + v) t

Show question

Question

You are provided u= Initial velocity, a= Acceleration, t= Time taken, and asked to find the final velocity. Which equation should you use?

v = u + at

Show question

Question

You are provided s= Displacement, v= Final velocity, t= Time taken, and asked to find acceleration. Which equation should you use?

s = vt - ½at²

Show question

Question

If the value of g is 10 m/s², how much time will it take for a ball (at rest) dropped from a large height to reach a velocity of 50 m/s.

v = u + at

=>  50 = 0 + 10t

=>  t = 5 seconds

Show question

Question

A car starts accelerating at 4 m/s² and crashes into a wall at 50 m/s after 5 seconds. How far away was the wall when the car started accelerating?

s = vt - ½at²

=>  s = 50 × 5 - ½ × 4 × 5² = 200

Show question

Question

A car is going around a circular race track with constant acceleration and a known initial velocity. Can you use SUVAT equations to find out displacement after a given amount of time?

No, SUVAT equations only apply to bodies moving in a straight line, and not in a circle.

Show question

Question

A footballer is performing high intensity training by running at full speed for a certain amount of time, then slowing down for a certain amount of time, and then going again at full speed for a certain amount of time. Can you use SUVAT equations to describe his motion over the entire course of his training?

No, because the acceleration is not constant.

Show question

Question

A driver applies the brakes and the car goes from 15 m/s to a halt within 5 seconds. How much distance did it travel before coming to a halt?

s = ½ (u + v) t

=>  s = ½ (15 + 0) 5 = 37.5 m

Show question

Question

A cyclist is traveling along a straight road. She accelerates at a constant rate from a velocity of 4 m/s to a velocity of 7 m/s² in 30 seconds. Find the distance she travels in these 30 seconds.

s = ½ (u + v) t

=> s = ½ × (4 + 7) × 30

= 165m

Show question

Question

A particle traveling at 12 m/s decelerates at a constant rate of 1.5 m/s² until it comes to a halt. For how long does the particle travel?

v = u +at

=> 0 = 12 - 1.5t

=> t = 8 seconds

Show question

Question

A particle traveling at 12 m/s decelerates at a constant rate of 1.5 m/s² until it comes to a halt. What distance does the particle travel?

v² = u² + 2 as

=> 0² = 12² - 2 × 1.5 × s

=> 3s = 144

=> s = 48 m

Show question

Question

A ball is thrown up with an initial velocity of 40 m/s. How long will it take the ball to reach its peak height assuming g = 10 m/s² ?

v = u + at

=> 0 = 40 - 10t

=> t = 4 seconds

Show question

Question

What is constant acceleration?

Acceleration is defined as the change in velocity over time. If the rate of change of velocity of a body remains constant over time, it is known as constant acceleration.

Show question

Question

Give an example of constant acceleration.

A body falling freely under the force of gravity with no other external force acting on it

Show question

Question

How do you find the velocity of an object in a straight line displacement-time graph?

Calculate the slope of the line

Show question

Question

A displacement-time graph shows a straight line parallel to the time axis. What is the acceleration?

Zero

Show question

Question

A displacement-time graph shows a straight line parallel to the time axis. What is the average velocity?

Zero

Show question

Question

A velocity-time graph shows a straight line parallel to the time axis. What is the acceleration?

Zero

Show question

Question

A velocity-time graph shows a straight line. What is the acceleration?

The acceleration is equal to the gradient of the line

Show question

Question

A velocity-time graph is curved and frequently changing direction. How do you find the acceleration at a given point in time?

Calculate the gradient of the graph at that point

Show question

Question

What are the five different SUVAT variables?

s= Displacement, u= Initial velocity, v= Final velocity, a= Acceleration, t= Time taken

Show question

Question

Usain is running around a circular race track with constant acceleration and a known initial velocity. Can you use SUVAT equations to find out his displacement after a given amount of time?

No, SUVAT equations only apply to bodies moving in a straight line, and not in a circle.

Show question

Question

You are provided u = Initial velocity, v = Final velocity, t = Time taken and asked to find displacement. Which equation should you use?

s = ½ (u + v) t

Show question

Question

What is the value of acceleration due to gravity?

Approximately 9.8 m/s²

Show question

Question

How much time will it take for a ball (at rest) dropped from a large height to reach a velocity of 29.4 m/s?

v = u + at

=>  29.4 = 0 + 9.8t

=>  t = 3 seconds

Show question

Question

A ball is thrown up with an initial velocity of 20 m/s. How long will it take the ball to reach its peak height? Assume g=10 m/s².

v = u + at

=> 0 = 20 - 10t

=> t = 2 seconds

Show question

Question

A ball is thrown up with an initial velocity of 20 m/s. What will be the peak height it reaches? Assume g=10 m/s².

v² = u² + 2 as

=> 0 = 20² - 2 × 10 × s

=> s = 20 m

Show question

Question

A cat and an elephant jump from the roof of a building at the same time. Who will hit the ground earlier? (Ignore the effect of air resistance.)

Both will hit the ground at the same time as the acceleration due to gravity does not depend on the mass of the falling object.

Show question

Question

What type of graph is used to display motion?

Two most commonly used graphs to display motion are displacement-time graphs and velocity-time graphs.

Show question

Question

An object is moving in a straight line with a constant acceleration of 5 m/s². Which of the following curves express it on a displacement-time graph?

Horizontal straight line

Show question

Question

An object is moving in a straight line with a constant acceleration of 5 m/s². Which of the following curves express it on a velocity-time graph?

Horizontal straight line

Show question

Question

An object is moving in a straight line with a constant velocity. Which of the following curves express it on a displacement-time graph?

Horizontal straight line

Show question

Question

An object is moving in a straight line with a constant velocity. Which of the following curves express it on a velocity-time graph?

Horizontal straight line

Show question

Question

An object is stationary. Which of the following curves express it on a velocity-time graph?

Horizontal straight line

Show question

Question

How do you calculate the total distance travelled from a given velocity-time graph?

Calculate the area enclosed between the graph and the time axis

Show question

Question

Which quantity can we obtain by calculating the gradient on a displacement-time graph?

acceleration

Show question

Question

Which quantity can we obtain by calculating the gradient on a velocity-time graph?

Acceleration

Show question

Question

A car is shown to accelerate from 0 m/s to 25 m/s in 80 seconds in a straight line on a velocity-time graph. What is the distance travelled in by the car in that time?

Distance travelled = area enclosed by the graph = base × height / 2 = 80 × 25 /2 = 1000 m.

Show question

Question

A car is shown to travel with a constant velocity of 13 m/s for 40 seconds on a velocity-time graph. What is the distance travelled in by the car in that time?

Distance travelled = area enclosed by the graph = 40 × 13 = 520 m.

Show question

Question

A runner is running around a circular race track at a speed of 7 m/s. Which of the following will be a straight line graph?

Displacement-time graph

Show question

Question

A man gets into his car, but the car refuses to start. What will be the resultant velocity-time graph depicting the motion of the car?

A straight line coinciding with the time axis (velocity = 0)

Show question

Question

A man gets into his car, but the car refuses to start. What will be the resultant displacement-time graph depicting the motion of the car?

A straight line coinciding with the time axis (displacement = 0)

Show question

Question

A velocity-time graph shows a straight line parallel to the time axis. What is the acceleration?

Zero

Show question

Question

A displacement-time graph shows a straight line parallel to the time axis. What is the acceleration?

Zero

Show question

Question

A displacement-time graph shows a straight line parallel to the time axis. What is the average velocity?

Zero

Show question

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