Question

A car moving with a velocity of 90 km/h decelerates uniformly at the rate of 2.5 m/s2
and
finally comes to rest. Find the time taken by the car to come to rest and the distance
covered by it during this time.

Answer 1

Given :-

Initial velocity = 90 km/h

Acceleration = -2.5 m/s

To Find :-

Time taken to stop

Distance covered

Solution :-

We know that

1 km/h = 5/18 m/s

90 km/h = 90 × 5/18 = 5 × 5 = 25 m/s

v = u + at

0 = 25 + (-2.5)(t)

0 - 25 = -2.5t

-25 = -2.5t

-25/-2.5 = t

25/2.5 = t

10 = t

v² - u² = 2as

(0)² - (25)² = 2(-2.5)(s)

0 - 625 = -5s

-625 = -5s

-625/-5 = s

625/5 = s

125 = s

Hence

Time taken is 10 second

Distance covered is 125 m