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:

Displacement-time graphs

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.

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.

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:

- s = displacement
- u = initial velocity
- v = final velocity
- a = acceleration
- 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.

- \(v = u + at\)
- \(s = \frac{1}{2} (u + v) t\)
- \(s = ut + \frac{1}{2}at^2\)
- \(s = vt - \frac{1}{2}at^2\)
- \(v^2 = u^2 + 2 as\)

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.

\(s = vt - \frac{1}{2}at^2\)

Solving for s you get:

\(s = 40 \cdot 5 - \frac{1}{2} \cdot 4 \cdot 5^2 = 150 m\)

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.

\(s = \frac{1}{2} (u + v) t\)

Solving for s:

\(s = \frac{1}{2} (15 + 0) 5 = 37.5 m\)

## 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.

\(s = ut + \frac{1}{2}at^2\)

Substituting all values to solve for t:

\(2.45 = 0 \cdot t + |frac{1}{2} \cdot 9.8 \cdot t^2\)

\(2.45 = 4.9t^2\)

\(t = \frac{1} {\sqrt 2} = 0.71 s\)

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

\(t = 2.6 s\)

## 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.

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##### Frequently Asked Questions about Constant Acceleration

Is acceleration due to gravity constant?

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.

What is constant acceleration in physics?

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.

How do you calculate 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.

What is the difference between constant velocity and acceleration?

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

What is the constant acceleration formula?

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.

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