Question

a car moving at the speed of 72 km/h is brought to rest in 10 sec by applying brakes . find the magnitude of average retardation due to brakes and distance travelled by the car after applying brakes

Answer 1

intial 72X1000/60X60 = 20 m/s
final 0
t =. 10 s
a = v-u
-------
t
= 0 - 20/10
= -2m/s sq

Answer 2

Answer:

The retardation is 2m/s² and the distance traveled by a car after applying brakes is 100m.

Explanation:

We will use two equations to solve this question i.e.

$v=u+at$     (1)

$s=ut+\frac{1}{2}at^2$ (2)

Where,

v=final velocity

u=initial velocity

a=acceleration

t=time

s=distance travelled

From question,

u=72km/hr=72×5/18=20m/s

v=0     (as the brake is applied)

t=10 sec

From equation (1),

$0=20+a\times 10$

$a=\frac{-20}{10}=-2m/s^2$     (negative sign shows that the acceleration is decreasing i.e.retardation)

Now by substituting the value of acceleration in equation (2) we get,

$s=20\times 10-\frac{1}{2}\times 2\times 10^2=100m$

Hence, the retardation is 2m/s² and the distance traveled by a car after applying brakes is 100m.