Question

a car moving alone a straight highway with a speed of 72 km hour sport to stop within a distance of 100 M what is the retardation of the car and how long does it take for the car to stop

Answer 1

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:
v2 – u2 = 2as
(0)2 – (35)2 = 2 × a × 200
a = – 35 × 35 / 2 × 200 = – 3.06 ms-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 s
Hope this helps you
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Answer 2

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V= 72km/h

v=0

s =100m

v^2-u^2= 2as

-u^2= 2×a×100(v=0)

- 20^2=200a

-400=200a

a=-2 m/s^2

time taken be t

a=(v-u)/t

or,

t=20/2=10s

I hope, this will help you

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