Motion — Question 2

Given,

Initial velocity (u) = 90 kmh^-1

Convert kmh^-1 to ms^-1

90 x $\dfrac{5}{18}$= 25 ms^-1

Terminal velocity (v) = 0

Acceleration (a) = -0.5 ms^-2

According to the third equation of motion,

v² – u² = 2as

$\Rightarrow \text{s} = \dfrac{\text{v}^2 - \text{u}^2}{2\text{a}}$

Substituting we get,

s = $\dfrac{0 - {25}^2 }{2 \times 0.5} = \dfrac{625}{1}$ = 625 m.

Hence, the train travels 625 m before it is brought to rest