Question

A car is moving in a straight line with speed 18 km h^-1. It is stopped in 5 s by applying the brakes. Find:

1. the speed of car in m s^-1
2. the retardation and
3. the speed of car after 2 s of applying the brakes.

Answer 1

Given, speed of car = 18 km h^-1

(i) To convert speed to m s^-1 

`18 " km h"^-1 = (18 " km")/(1 " h") = (18 xx 1000 " m")/(60 xx 60 " s")`

`=> 18 " km h"^-1 = (18 xx 10 " m")/(6 xx 6 " s")`

`=> 18 " km h"^-1 = 5 "m s"^-1`

Hence, 18 km h^-1 is equal to 5 m s^-1.

(ii) As the car is stopped, the final velocity = 0

initial velocity (u) = 18 km h^-1 

As we know,

Acceleration (a) = `("Final velocity (v)" - "initial velocity (u)")/("time (t)")`

Given, Initial velocity (u) = 5 m s^-1

Final velocity (v) = 0

Time (t) = 5 s

Substituting the values in the formula above, we get,

Acceleration (a) = `(0 - 5)/5`

Acceleration (a) = `- 5/5`

Acceleration (a) = - 1 ms^-2

Hence, acceleration of the car is -1 m s^-2. Negative sign shows that the velocity decreases with time, so retardation is 1 m s ^-2.

(iii) Given, t = 2 s, a = -1 m s^-2, u = 5 m s^-1

Substituting the values in the formula for acceleration, we get,

`=> - 1 "ms"^-2 = ((v - 5)"ms"^-1)/(2s)`

`=> -2ms^-1 = (v - 5)ms^-1`

⇒ v = (5 - 2) ms^-1

⇒ v = 3 ms^-1

Hence, the speed of car after 2 s of applying the brakes = 3 m s^-1.