Question

Q.4 - Starting from stationary position, Rahul paddles his bicycle to attain a velocity of 6m/s in 30 minutes. Then he applies brakes such that the velocity of bicycle comes down to 4m /s in next 5 S. Calculate the acceleration of bicycle in both cases

Answer 1

Answer:

Explanation:

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.

So from equation (1) substitute the values we have,

⇒6=0+a(30)

⇒a=630=0.2ms−2.

So in the first case his acceleration is 0.2ms−2 .

Now in the second case he applies brakes such that the velocity of the bicycle comes down to 4m/s in the next 5s.

So his initial velocity becomes 6 m/s and final velocity becomes 4 m/s and the time taken is 5s.

So substitute these values in equation (1) we have,

⇒4=6+a(5)

⇒a=4−65=−25=−0.4ms−2 (‘-’ sign indicates retardation).

So this is the required answer.

Note – The key concept here was the negative sign of acceleration, retardation is opposite of acceleration and it generally comes into play when the final velocity of an object is less than the initial velocity of the object in a path. Acceleration is itself a vector quantity as it has both direction as well as magnitude.

Answer 2

Given : Starting from Rest , Rahul paddles his bicycle to attain a velocity of 6m/s in 30 minutes

Applies brakes such that the velocity of bicycle comes down to 4m /s in next 5 secs

To Find :  the acceleration of bicycle in both cases

Solution:

Starting from stationary position

Hence u = 0  

Attain a velocity of  6 m/s in 30 minutes

v = 6 m/s

t = 30 mins = 1800 sec

a = (v - u)/t

=> a = (6 - 0)/1800

=> a = 1/300  m/s²

u = 6 m/s

v = 4 m/s

t = 5 sec

a = ( 4 - 6)/5  = - 0.4 m/s²

Acceleration in 1st case is  1/300  m/s²

Acceleration  in 2nd case is  - 0.4 m/s²

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