Question

A motorbike moving with initial velocity 28 m s^–1 and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.

Answer 1

Given,

Initial velocity (u) = 28 m s^–1

Final velocity (v) = 0 (comes to a stop)

Distance travelled (s) = 98 m

According to the third kinematic equation,

v² = u² + 2as

Substituting, we get,

(0)² = (28)² + 2 × a × 98

0 = 784 + 196a

⇒ a = `(-784)/196 -4` m s^–2

Hence, the acceleration of the motorbike is −4 m s^–2.

The motorcycle is slowing down when the acceleration is opposite to the direction of motion, as indicated by the negative sign.

Using the first kinematic equation for the duration,

v = u + at

or

t = `(v - u)/a`

Substituting, we get,

t = `(0 - 28)/-4 = 7` s

Hence, the time taken to come to a stop is 7 s.