A car travelling with a speed 126kmph along a straight line comes to rest after travelling a distance 245m.The time taken by the car to come to rest,in second is
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Initial velocity of the car, u=126km/h = 35m/s
Final velocity of the car, v=0
Distance covered by the car before coming to rest, s =245m
Retardation produced in the car= a
From third equation of motion, a can be calculated as:
v^2-u^2=2as
0-(35)^2=2*a*245
a=-(35*35)/(2*245)
=-(1225)/490
=-2.5 m/s^2
From first equation of motion, time (t) taken by the car to stop can be obtained as:
v=u+at
or t=-35/-2.5=14
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