Question

A car starts from rest and travels 5.0 s with a uniform acceleration of 1.5 m/s2. The driver then applies the brakes, causing the car to slow down at a rate of 2.0 m/s2. If the brakes are applied for 3.0 seconds, how fast is the car going at the end of the braking period and how far has it gone from the start?

Answer 1

1.5 m/s

Step-by-step explanation:

Answer 2

1.5 m/s

32.25 m

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

a = Acceleration

s = Displacement

Equation of motion

`v=u+at\\\Rightarrow v=0+1.5* 5\\\Rightarrow v=7.5\ m/s`

The velocity of the car at the end of the acceleration period is 7.5 m/s

This velocity will be considered as the initial velocity of the decelerating stage

`v=u+at\\\Rightarrow v=7.5-2* 3\\\Rightarrow v=1.5\ m/s`

The velocity of the car at the end of the deceleration period is 1.5 m/s

`s=ut+(1)/(2)at^2\\\Rightarrow s=0* t+(1)/(2)* 1.5* 5^2\\\Rightarrow s=18.75\ m`

Distance traveled in the acceleration period is 18.75 m

`s=ut+(1)/(2)at^2\\\Rightarrow s=7.5* 3+(1)/(2)* -2* 3^2\\\Rightarrow s=13.5\ m`

Distance traveled in the deceleration period is 13.5 m

Total distance traveled is 18.75+13.5 = 32.25 m