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Constant Acceleration Equations

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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:

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

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

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

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

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

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.

v = u + at

s = ½ (u + v) t

s = ut + ½at²

s = vt - ½at²

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.

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

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

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.

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

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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