Question

a car moving along a straight highway with speed of 126 km h power -1 is brought to a stop within a distance 200 m. what is the retardation of the car (assumed uniform ) and how long it take for the car to stop?

Answer 1

v^2-u^2=2as

u=126km/hr
=126×5/18
=35
v=0
s=200m
a=?
0^2- 35^2=2×a×200
0-1225=2×a×200
-1225=400a
a=-1225/400=-3.06

v=u+at
0=35-3.06×t
3.06t=35
t=35/3.06
t=11.43s

Answer 2

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Initial velocity of the car, u = 126 km/h = 35 m/s

Final velocity of the car, v = 0

Distance covered by the car before coming to rest, s = 200 m

Retardation produced in the car = a

From third equation of motion, a can be calculated as:

V^2 - U^2 = 2aS

(0)^2 - (35)^2 = 2 . a . 200

a = ( 35×35)/(2× 200)

a= 3.06 m/ s^2

From first equation of motion, time (t) taken by the car to stop can be obtained as:

V = U + at

t = (V- U )/a = (-35)/(-3.06) = 11.44 sec

I hope, this will help you

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