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# Constant Acceleration Equations

For a body moving in a single direction with constant acceleration, the constant acceleration or SUVAT equations are used to connect five different variables of motion. These variables are:

1. s = Displacement – the total displacement of the body since the beginning of the measure at a given point of time.

2. u = Initial velocity – the velocity of the body at the beginning of the measurement.

3. v = Final velocity – the velocity of the body at the end of the measurement.

4. a = Acceleration – the constant acceleration of the object throughout the measurement.

5. t = Time taken – the time elapsed from the start to the end of the measurement.

## The five constant acceleration equations

There are five different constant acceleration equations that are used to connect and solve for the variables above. It is a good idea to learn these equations by heart.

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. Given any three variables, it would be possible to solve for any of the other two variables.

When can you use the SUVAT equations? The SUVAT equations apply for a body moving in a straight line with constant acceleration.

## Deriving the constant acceleration equations

Let us look at how we obtain these equations.

Equation 1: By definition, acceleration is the change in velocity per unit time. The following diagram demonstrates the concept.

A body with initial velocity u accelerates with a constant acceleration to attain a final velocity v after time t.

Let us express the definition mathematically.

acceleration = (change in velocity)/(change in time)

=> a = (v - u)/t

Rearranging the above equation, we get the first equation :

v = u + at

Equation 2: Remember, we are dealing with constant acceleration here. So the average velocity across the duration of motion is (u + v)/2. Multiplying the average velocity by time gives the displacement. Therefore,

s = (u + v)/2 × t

This gives us the second equation,

s = ½ (u + v) t

Equation 3: To obtain the third equation, directly substitute the value of v from the first equation into the second equation.

s = ½ (u + v) t

=> s = ½ (u + u + at) t

This gives us the third equation,

s = ut + ½at²

Equation 4: To obtain the fourth equation, first rearrange the first equation and express it in terms of u.

=> u = v – at

Substitute this value of u into the third equation,

s = ut + ½at²

=> s = (v - at) t + ½at²

This gives us the fourth equation,

s = vt - ½at²

Equation 5: To obtain the fifth equation, first rearrange the first equation and express it in terms of t.

=> t = (v - u)/a

Substitute this value of t into the second equation,

s = ½ (u + v) t

=> s = ½ (u + v) (v - u)/a

=> 2as = (v + u) (v – u)

=> 2as = v² - u²

This gives us the fifth equation,

v² = u² + 2 as

## Solving problems using constant acceleration equations

Let us look at some examples of problems that can be solved using the constant acceleration equations.

A car with an initial velocity of 8 m/s is accelerating at a rate of 2 m/s². How long will it take to reach a speed of 20 m/s?

Solution 1

Here, v = 20 m/s, u = 8 m/s, a = 2 m/s². v = u + at

=> t = (v - u)/a

=> t = (20 - 8)/2

= 6 seconds

A car with an initial velocity of 8 m/s is accelerating at a rate of 2 m/s². How long will it take to travel a distance of 200m?

Solution 2

Here, s = 200 m, u = 8 m/s, a = 2 m/s².

s = ut + ½at²

=> 65 = 8t + ½ × 2t²

=> t² + 8t – 65 = 0

=> t = 5

Note: The obtained quadratic equation gives two values, 5 and -13. Time cannot be negative, so we take the positive value as the answer.

A marathon runner decides to accelerate during the last 200 meters of a race. He accelerated at a rate of 0.07 m/s², eventually crossing the finish line at a speed of 8 m/s. What speed was he running at before he decided to accelerate?

Solution 3

Here, s = 200 m, v = 8 m/s, a = 0.07 m/s².

v² = u² + 2 as

=> u² = v² - 2as = 8 × 8 - 2 × 0.07 × 200

=> u² = 36

=> u = 6 m/s

A cyclist is travelling along a straight road. It accelerates at a constant rate from a velocity of 4 m/s to a velocity of 7.5 m/s² in 40 seconds. Find a) the distance she travels in these 40 seconds. b) her acceleration in these 40 seconds.

Solution 4

a) s = ½ (u + v) t

=> s = ½ × (4 + 7.5) × 40 = 230m

b) v = u + at

=> 7.5 = 4 + 40a

=> a = (7.5 - 4)/40 = 0.0875 m/s²

A ball is thrown up with an initial velocity of 39.2 m/s. How long will it take the ball to reach its peak height assuming g = 9.8 m/s²

Solution 5

Here, the acceleration of the ball is -9.8 m/s², as it is the force of gravity that is slowing the ball down.

v = u + at

=> 0 = 39.2 - 9.8t

=> t = 4 seconds

## Constant Acceleration Equations - Key takeaways

• The constant acceleration equations are used to connect five different variables: s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time taken.

• The constant acceleration equations are applicable for a body moving in a straight line with constant acceleration.

• The constant acceleration equations can be derived starting from the basic definition that acceleration is the change in velocity per unit time.

• Each constant acceleration equation contains four of the five SUVAT variables. Given any of the three variables of an equation, it should be possible to solve for the fourth variable as the unknown.

The equation for constant acceleration is v = u + at, where u= Initial velocity, v= Final velocity, a= Acceleration, t= Time taken

There are 5 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

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.

An example of constant acceleration is a body falling under the force of gravity with no other external force acting on it. In reality, it is very difficult to achieve perfect constant acceleration because there are typically multiple forces acting on any object.

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 Equations 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

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