1.

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

Answer»

SOLUTION :In the first case:
initial velocity `u = 0,`
FINAL velocity `v = 6 MS ^(-1),`
time `t = 30 s.`
`a = (v - u)/(t)`
Subsituting the given values of u, v and t in the above EQUATION, we get
`a = (( 6 ms ^(-1) -0 ms ^(-1)))/(30S)= 0.2 ms ^(-2)`
In the second case :
In the second case:
initial velocity `u = 6 ms ^(-1),`
final velocity `v = 4 ms ^(-1),`
time `t =5s.`
Then,` a = (( 4 ms ^(-1) - 6 ms ^(-1)))/(5s) =-0.4 ms ^(-2)`
The acceleration of the bicycle in the first case is `0.2 ms ^(-2)` and in the second case, it is `0.4 ms ^(-2).`


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