Question

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

1. The initial velocity of the bus
2. The distance travelled by bus during this time.

Answer 1

Final velocity (v) = 0

retardation = -25 cm/s² 

Expressing it in m s^-2 

25 cm s^-2 = `25/100` m s^-2 

Hence, - a = - 0.25 m s^-2 

Time taken (t)= 20 s

(i) Let 'u' be the initial velocity.

Using the first equation of motion,

v = u + at

We get,

u = v - at

u = 0 - (- 0.25)(20)

u = 5 m s^-1

(ii) Let 's' be the distance travelled.

Using the third equation of motion,

v² - u² = 2as

We get,

∴ (0) ² - (5)² = 2 (-0.25) (s)

∴ 0 - 25 = - 0.5 × s

∴ - 25 = - 0.5 s

∴ `(- 25)/(- 0.5)` = s

∴ s = 50 m.