Question

A car starts from rest and travels for 9.0 s with a uniform acceleration of +2.4 m/s²?. The driver then applies the brakes, causing a uniform acceleration of -2.5 m/s². If the brakes areapplied for 2.0 s, determine each of the following(a) How fast is the car going at the end of the braking period?m/s(b) How far has the car gone?m

Answer 1

Answer:

a) At the end of the braking period, the car is moving at a speed of 16.6m/s

b) The car has traveled a total distance of 135.4m

Explanations:

The car starts from rest

The initial velocity, u = 0 m/s

Uniform acceleration, a = 2.4 m/s²

time, t = 9.0 seconds

Find the final velocity when the car accelerates at 2.4m/s² using the equation

v = u + at

v = 0 + 2.4(9)

v = 21.6 m/s

The distance covered when the car accelerates at 2.4m/s²

s = ut + 0.5at²

s = 0(9) + 0.5(2.4)(9²)

s = 97.2 m

The distance covered when the car accelerates at 2.4m/s² is 97.2 m

The driver then applied a brake for 2.0 s and accelerates at -2.5m/s²

The initial velocity now becomes 21.6 m/s

That is, u = 21.6 m/s

t = 2 seconds

a = -2.5m/s²

The final speed v is calculated as:

v = u + at

v = 21.6 + (-2.5)(2)

v = 21.6 - 5

v = 16.6m/s

At the end of the braking period, the car is moving at a speed of 16.6m/s

The distance covered during the braking period is calculated as:

`\begin{gathered} s\text{ = (}(u+v)/(2))t \\ s\text{ = }(21.6+16.6)/(2)*2 \\ s\text{ = }(38.2)/(2)*2 \\ s\text{ = 38.2 m} \end{gathered}`

The car traveled a distance of 38.2 m during the braking period

Total distance covered = 97.2m + 38.2m

Total distance covered = 135.4m

The car has gone a distance of 135.4m