Question

A car moving along a straight road with a
speed of 120 km/hr, is brought to rest by
applying brakes. The car covers a distance
of 100 m before it stops. Calculate (i) the
average retardation of the car (ii) time
taken by the car to come to rest.​

Answer 1

(i)72 m/s^2 is the answer. formula :v^2=u^2-2as

(ii)formula : v =u-at.

ans,t=u/a=5/3=1.67 second

Answer 2

Answer:

(i)$a=-72000 km/hr^2$

(ii) 6 sec

Explanation:

We are given that  a car moving along a straight road with a  speed 120 km/hr.

Final velocity=0 km/hr

Initial velocity=120 km/hr

Distance covered by the car=100 m=$\frac{100}{1000}=0.1 Km$

We have to (i) the average retardation of the car

(ii) time taken by the car to come to rest

We know that

$v^2-u^2=2as$

Substitute the value then we get

$0-(120)^2=2\times\frac{1}{10}a$

$a=-14400\times 5 km/hr^2$

$a=-72000 km/hr^2$

Retardation is an acceleration in opposite direction.

$-72000 =\frac{v-u}{t}$ substitute the value then we get

$72000=\frac{-120}{t}$

$t=\frac{120}{72000}$

$t=0.00167 hr$

$t=0.00167\times 3600=6 sec$

Because 1 hr=3600 sec

Hence, the taken by the car to come to rest =6 sec.