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Question
A motor bike running at 90 kmh−1 is slowed down to 18 kmh−1 in 2.5 s. Calculate
- acceleration
- distance covered during slow down.
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Solution
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
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