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

(i) From A to B:
v² − u² = 2a S
(15)² − (25)² = 2a (40)
225 − 625 = 80a
80a = −400
a = −5 ms⁻²

Time (t₁) = ?
v = u + at₁
15 = 25 + (−5) t₁
5t₁ = 25 − 15 = 10
t₁ = `10/5` = 2s

From B to C:
Initial velocity = u = 54 kmh⁻¹ = `54xx5/15` ms⁻¹
u = 18 ms⁻¹
Final velocity = v = 0
Acceleration = a = −5 ms⁻²
Time = t₂ = ?
v = u + at₂
0 = 18 + (−5) t₂
5t₂ = 18
t₂ = `18/5` = 3.6 s

(ii) Distance covered from B to C:
u = 54 kmh⁻¹ = 18 ms⁻¹
v = 0
a = −5 ms⁻²
S = ?
v² − u² = 2a S
(0)² − (18)² = 2(−5)S
−10S = −324

S = `324/10` = 32.4 m
Total distance covered from A to C = 40 + 32.4 = 72.4
Total time taken from A to C = t₁ + t₂ + = 2 + 3.6 = 5.6s