Question

A car moving along a straight highway with speed of 72 km/h-1 is brought to a stop within distance 100 m.what is retardation of the car and how long it take for car to stop solve it

Answer 1

initial velocity (u)=72km/h=72×5/18=20m/s
final velocity (v)=0, distance (s)=100m ,a be the acceleration,t be the time
v×v-u×u=2(-a)s
0-20×20=-2a×100
a=2m/s^2
v=u+(-a)t
0=20-2t
t=10seconds
retardation=2m/s^2
time=10seconds

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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