Question

A car of mass 2400 kg moving with a velocity of 20 m/s is stopped in 10 seconds by applying brakes calculate the retardation and the retardating force

Answer 1

Explanation:

u= 20 m/s

v= 0m/s

t= 10 s

a= v-u / t = 0 -20/10= -2m/s is the retardation . it is negative as acc is in opp direction of cars motion

F = 2400* -2 N= -4800N. negative as force is applied in the opp direction to the direction of motion of the car

Answer 2

Answer:-

a = -2 m/s²

F = -4800 N

Given :-

m = 2400 kg

u = 20 m/s

v = 0 m/s

t = 10 s

To find :-

The retardation produced and the magnitude of retarding force.

Solution:-

After applying the brakes car will come to rest and v = 0 m/s

Now,

We have to find retardation produced:-

$\mathsf{a = \dfrac{v-u}{t}}$

$\mathsf{ a = \dfrac{0 -20}{10}}$

$\mathsf{a = \dfrac{-20}{10}}$

$\mathsf{a = -2 m/s^2 }$

hence,

Retardation produced in wheel be -2 m/s².

Now,

Magnitude of retardation force is given by using 2nd law of Newton.

→$\mathsf{ F = ma }$

→$\mathsf{F = 2400 \times -2}$

→$\mathsf{F = -4800 N}$

hence,

Magnitude of Retardation force will be -4800 N.