A car moving along a straight highway with speed of 126 km h-1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ?
Answers
Answered by
6
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
Answered by
2
Answer:
Hey mate ur answer is 39360km/h²
Explanation:
Initally speed of car=126km/hr
final speed of car=0km/hr
distance covered by the car=.2km
By the third equation of motion we have
v²=u²+2as
0=(126)²+2×0.2×a
0=15876+0.4a
0.4a=-15876
4a=-158760
a=-39690km/h²
retardation=39690km/h²
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