A car moving along a straight highway with speed of 126km/h is brought to stop with a distance of 200 m what is the retardation of the car and how long does it take for the car to stop?
Answers
Answered by
3
Explanation:
u = 126km/h = 35m/s
v = 0 m/s
s = 200m
using v*2 = u*2 + 2as
0 = 35*2 - 2a × 200
negative sign due to retardation
1225 = 400a
calculating this you will get
a = 3m/ s*2
Answered by
1
Answer:
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
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