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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
- total time in which bike comes to rest
- total distance travelled by bike.
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Solution
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
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