Question

A motor bike running at 90 kmh⁻¹ is slowed down to 18 kmh⁻¹ in 2.5 s. Calculate

1. acceleration
2. distance covered during slow down.

Answer 1

Initial velocity of motor bike = u = 90 kmh⁻¹
= u = `90xx5/18` ms⁻¹ = 25 ms⁻¹
Final velocity of motor bike = v = 18 kmh⁻¹
v = `18xx5/18` ms⁻¹ = 5 ms⁻¹
Time = t = 2.5 s

(i) Acceleration = a = ?
v = u + at
5 = 25 + a (2.5)
2.5a = −25 + 5 = −20
a = `(-20)/2.5` = −8 ms⁻²

(ii) Distance covered S =?
v² − u² = 2aS
(5)² − (25)² = 2(−8)S
25 − 625 = −16S
−16S = −600
S = `600/16`
S = 37.5 m