Motion in One Dimension — Question 8

(i) As we know, according to the equation of motion,

v = u + at

Given,

u = 90 km h^-1

velocity in m s^-1

$90 \text{km h}^{-1} = \dfrac{90 \text{km}}{1\text{h}} = \dfrac {90 \times 1000 \text{m}}{60 \times 60 \text{s}} \\[0.5em] \Rightarrow 90 \text{km h}^{-1} = \dfrac {90 \times 10 \text{m}}{6 \times 6 \text{s}} \\[0.5em] \Rightarrow90 \text{km h}^{-1} = \dfrac {15 \times 10 \text{m}}{1 \times 6 \text{s}} \\[0.5em] \Rightarrow 90 \text{km h}^{-1} = \dfrac {25 \times \text{m}}{\text{s}} \\[0.5em]$

Hence, 90 km h^-1 is equal to 25 m s^-1.

v = 0 (as the train stops on application of brakes)

t = 10 s

retardation = -a = -0.5 m s^-2

Substituting the values in the formula above we get,

$\text {v} = 25 + [(-0.5) \times 10] \\[0.5em] \text {v} = 25 + [- 5] \\[0.5em] \Rightarrow \text {v} = 20 \text {m s}^{-1} \\[0.5em]$

Hence, the velocity after 10 s = 20 m s^-1.

(ii) As we know, according to the equation of motion,

v = u + at

Substituting the values in the formula above, to get the time taken by the train to come to rest.

$\text {0} = 25 + (-0.5 \times \text{t}) \\[0.5em] \text {0} = 25 - 0.5 \text{t} \\[0.5em] 0.5 \text{t} = 25 \\[0.5em] \Rightarrow \text {t} = \dfrac {25}{0.5} \\[0.5em] \Rightarrow \text {t} = \dfrac {250}{5} \\[0.5em] \Rightarrow \text {t} = 50\ \text {s} \\[0.5em]$

Hence, the time taken by the train to come to rest = 50 s.