Question

A motorbike running at 90 kmh⁻¹, is slowed down to 54 kmh⁻¹ by the application of brakes, over a distance of 40 m. If the brakes are applied with the same force, calculate

1. total time in which bike comes to rest
2. total distance travelled by bike.

Answer 1

Initial velocity = u = 90 kmh⁻¹
= u = `90xx5/18` ms⁻¹ = 25 ms⁻¹
Final velocity = v = 54 kmh⁻¹ = `54xx5/18` ms⁻¹
v = 15 ms⁻¹
Distance = S = 40 m

v² = u² + 2as

(15)² = (25)² + 2 ⋅ a ⋅ 40 ⇒ 225 = 625 + 80a ⇒ 80a = 225 − 625 = −400 ⇒ a = `(-400)/80`

a = −5 m/s²

v = u + at ⇒ 0 = 25 + (−5)t ⇒ t = `25/5` = 5 seconds

v2 = u2 + 2as ⇒ 0 = (25) 2 + 2 ⋅ (−5) ⋅ s ⇒ 625 = 10s ⇒ s = `625/10` = 62.5 m

Total time to stop = 5 seconds

Total distance before stopping = 62.5 m