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Question
When brakes are applied to a bus, retardation produced is 25 cm s-2 and the bus takes 20 s to stop. Calculate -
- The initial velocity of the bus
- The distance travelled by bus during this time.
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Solution
Final velocity (v) = 0
retardation = -25 cm/s2
Expressing it in m s-2
25 cm s-2 = `25/100` m s-2
Hence, - a = - 0.25 m s-2
Time taken (t)= 20 s
(i) Let 'u' be the initial velocity.
Using the first equation of motion,
v = u + at
We get,
u = v - at
u = 0 - (- 0.25)(20)
u = 5 m s-1
(ii) Let 's' be the distance travelled.
Using the third equation of motion,
v2 - u2 = 2as
We get,
∴ (0) 2 - (5)2 = 2 (-0.25) (s)
∴ 0 - 25 = - 0.5 × s
∴ - 25 = - 0.5 s
∴ `(- 25)/(- 0.5)` = s
∴ s = 50 m.
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