When brakes are applied to a bus, the retardation produced is 25 cm s^-2 and the bus takes 20 s to stop.

Calculate —

(i) the initial velocity of bus, and

(ii) the distance travelled by bus during this time.

As we know,

v = u + at

Given,

Retardation = -a = -25 cm s^-2

Expressing it in m s^-2

25 cm s^-2 = $\dfrac{25}{100}$m s^-2

Hence, -a = -0.25 m s^-2

t = 20 s

v = 0

u = ?

Substituting the values in the formula above we get,

$0 = u + (- 0.25) \times 20 \\[0.5em] 0 = u - 5.0 \\[0.5em] \Rightarrow u = 5 \text{ms}^{-1}$

Hence, the initial velocity of bus = 5 m s^-1

(ii) As we know, from the equation of motion,

v² = u² + 2aS

Substituting the values, we get,

$0^2 = 5^2 + [2 \times (- 0.25) \times S] \\[0.5em] 0 = 25 - [0.50 \times S] \\[0.5em] 0.50 \times S = 25 \\[0.5em] \Rightarrow S = \dfrac{25}{0.5} \\[0.5em] \Rightarrow S = \dfrac{250}{5} \\[0.5em] \Rightarrow S = 50 m \\[0.5em]$

Hence, distance travelled = 50 m