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Question
A cyclist driving at 5 ms−1, picks a velocity of 10 ms−1, over a distance of 50 m. Calculate
- acceleration
- time in which the cyclist picks up above velocity.
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Solution
Initial velocity = u = 5 ms−1
Final velocity = v = 10 ms−1
Distance = S =50 m
(i) v2 – u2 = 2aS
(10)2 – (5)2 = 2a (50)
100 − 25 = 100a
100a = 75
a = `75/100` = 0.75 ms−2
∴ Acceleration = a = 0.75 ms−2
(ii) v = u + at
10 = 5 + 0.75t
0.75t = 10 − 5 = 5
(Time) t = `5/0.75` = 6.67 S
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