A motorcycle starting from rest is uniformly accelerated such that its velocity in 10s is 72kmhr\(^{-1}\). What is its acceleration?
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Correct Answer: Option D
Explanation:
where u = o m/s, t = 10s,
v = 72kmhr\(^{-1}\) ---> (72 X 1000) / 3600
: v = 20 m/s
Acceleration (a) = ( v - u ) / t ---> (20 - 0) / 10
: a = 2m/s\(^2\)
where u = o m/s, t = 10s,
v = 72kmhr\(^{-1}\) ---> (72 X 1000) / 3600
: v = 20 m/s
Acceleration (a) = ( v - u ) / t ---> (20 - 0) / 10
: a = 2m/s\(^2\)