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

Find —

(i) the speed of car in m s ^-1,

(ii) the retardation and

(iii) the speed of car after 2 s of applying the brakes .

Given,

speed of car = 18 km h^-1

(i) To convert speed to m s^-1

$18 \text {km h}^{-1} = \dfrac{18 \text {km}}{1 \text{h}} = \dfrac {18 \times 1000 \text{m}}{60 \times 60 \text {s}} \\[0.5em] \Rightarrow 18 \text {km h}^{-1} = \dfrac {18 \times 10 \text {m}}{6 \times 6 \text {s}} \\[0.5em] \Rightarrow 18 \text {km h}^{-1} = 5 \text {m s}^{-1} \\[0.5em]$

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,

$\text {a} = \dfrac {\text {final vel. (v)} - \text {initial vel. (u)}}{\text {time (t)}}$

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,

$\text {Acceleration (a)} = \dfrac {0 - 5}{5} \\[0.5em] \text {Acceleration (a)} = -\dfrac {5}{5} \\[0.5em] \Rightarrow \text {Acceleration (a)} = - 1 \text {ms}^{-2} \\[0.5em]$

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,

$- 1 \text {ms}^{-2} = \dfrac {(v - 5 ) \text {ms}^{-1}}{2 \text {s}} \\[0.5em] - 2 \text {ms}^{-1} = (v - 5) {ms}^{-1} \\[0.5em] \Rightarrow v = (5 - 2) \text {ms}^{-1} \\[0.5em] \Rightarrow v = 3 \text {ms}^{-1} \\[0.5em]$

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