Question

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 bicycles come down to 4ms^-1 in the next 5s. Calculate the acceleration of the bicycle in both the cases ?

Answer 1

Answer:

Then he applies brakes such that the velocity of the bicycle comes down to 4ms−1 in the next 5 sec. ... Now in the first case, Rahul paddles from the starting point so his initial velocity is zero and attains a velocity of 6 m/s in 30 sec. Therefore his final velocity is 6 m/s and he takes time 30 sec.

Hope it helps

Answer 2

Given :

• 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 bicycles come down to 4ms^-1 in the next 5s.

To find :

• Calculate the acceleration of the bicycle in both the cases.

Solution :

In the first case :

u = 0

v = 6m/s-1

t = 30 s

We have :

$\: \: \: \: \: \: \: \: \: \: \: \: \rightarrow \sf \: a = \frac{v - u}{t}$

• Substituting the given values of u,v and t in the above equation.

we get :

$\boxed{ \sf \: a = \frac{(6m {s}^{ - 1} - 0ms {}^{ - 1} )}{30s} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \rightarrow \sf0.2ms {}^{ - 2}$

In the second case :

u = 6ms-1

v = 4ms-1

t = 5s

Then :

$\boxed{ \sf \: a = \frac{(4m {s}^{ - 1} - 6m {s}^{ - 1} )}{5s} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rightarrow \sf - 0.4m {s}^{ - 2}$

The acceleration of the bicycle in the first case is 0.2ms-2 and in the second case, it is -0.4ms-2.