Question

The mass of a cyclist together with a bicycle is 100 kg. He is driving at 10 m/s and stops his mount in 2s by the application of brakes. Calculate the i) retardation ii) distance covered during the action of application of brakes iii) retarding force

Answer 1

QUESTION :-

The mass of a cyclist together with a bicycle is 100 kg . He is driving at 10 m/s and stops his mount in 2s by the application of brakes . Calculate the

i )  retardation

ii ) Distance covered during the action of application of brakes

iii ) retarding force

SOLUTION :-

GIVEN -

mass of the cyclist together with the bicycle  ( m ) = 100 kg

initial velocity ( u ) = 10 m/s

Time ( t ) = 2 s

Final velocity ( v ) = 0 m/s

since, break is applied and the cycle finally stops , hence, the final velocity is equal to zero.

( i )  Retardation

First equation of motion

   v  =  u  +  at

where,

    v  = final velocity

    u = initial velocity

    a = acceleration

     t = time

putting the given values in the formula,

   0 m/s =  10  m/s  + a × 2 s

→   2a = - 10 m/s²

→   a  = - 5 m/s²

therefore,  

accleration = - 5 m/s²

thus,   retardation = 5 m/s²

( ii )  Distance covered during the action

Second Equation of Motion

   S = ut + 1/2 at²

where,

    S = distance covered

    u = initial velocity

     t = time

 

substituting the given values in the formula

    S  = 10 m/s ×  2 s   + 1/2 × ( -5m/s² ) × (2s)²

→   S =  20 m  -  10 m

→   S = 10 m

Thus,  

the distance covered during the action of application = 10 m

( iii )  Retarding force

       Retarding force = mass × retardation

→        Retarding force  =  100 kg × 5 m/s²

→       Retarding force = 500 kg m/s²

→       Retarding force = 500 N