Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of $6 m s^{- 1}$ in $30 s .$ Then he applies brakes such that the velocity of the bicycle cones down to $4 m s^{- 1}$ in the next $5 s .$ Calculate the acceleration of the bicycle in both the cases.

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Solution

For first case:
Initial velocity $= 0$
Final velocity $= 60 m / s$
Time $= 30 sec$
$v = u + a t$
$6 = 0 + a \times 30$
$6 = 30 a$
$a = \frac{6}{30} = 0.2$
$a = 0.2 m / s^{2}$
For second case:
Initial velocity $= 6 m / s$
Final velocity $= 4 m / s$
Time taken $= 5 sec$
$v = u + a t$
$4 = 6 + a \times 5$
$4 - 6 = 5 a$
$- 2 = 5 a$
$a = - \frac{2}{5}$
$a = - 0.4 m / s^{2}$

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Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6ms−1 in 30s. Then he applies brakes such that the velocity of the bicycle cones down to 4ms−1 in the next 5s. Calculate the acceleration of the bicycle in both the cases.

For first case:Initial velocity =0Final velocity =60m/sTime =30secv=u+at6=0+a×306=30aa=630=0.2a=0.2m/s2For second case:Initial velocity =6m/sFinal velocity =4m/sTime taken =5secv=u+at4=6+a×54−6=5a−2=5aa=−25a=−0.4m/s2

Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of $6 m s^{- 1}$ in $30 s .$ Then he applies brakes such that the velocity of the bicycle cones down to $4 m s^{- 1}$ in the next $5 s .$ Calculate the acceleration of the bicycle in both the cases.