A car is moving with a speed of 126 km/hr is brought to stop within a distance of 200 m claculate the retardation
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we can apply 3 eq^n of motion
v^2-u^2 =2as
v= 0m/s
u=126 km/h *5/18=35m/s
s=200m
0^2-35^2=2a*200
-1225=400a
a=-3.1 m/s^2
retardation is 3.1m/s^2
v^2-u^2 =2as
v= 0m/s
u=126 km/h *5/18=35m/s
s=200m
0^2-35^2=2a*200
-1225=400a
a=-3.1 m/s^2
retardation is 3.1m/s^2
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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
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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