Answer to Question #6947 in Mechanics | Relativity for MS MANOJ

Question #6947
A VEHICLE IS TRAVELLING AT 30KM/H WHEN ITS POWER IS SHUT OFF.IT TRAVELS 0.9KM BEFORE IT FINALLY COME TO REST.CALCULATE THE TIME TAKEN FOR THE VEHICLES TO COME TO REST AND THE VALUE OF THE RETARDATION.
1
Expert's answer
2012-03-01T10:34:53-0500
We'll use the next formula:
V(t) = at + V0,
where V0 = 30KM/H and a is the retardation.
Let's T be the time when the vehicle stops. Then
V(T) = 0 and aT + V0 = 0 ==> a = -V0/T.
Also we know that
S = 0.9km = aT²/2 + V0*T, so
aT²/2 + V0*T = 0.9, or
(-V0/T)*T²/2 + V0*T = 0.9, or
-V0*T/2 + V0*T = 0.9, or
V0*T/2 = 0.9 ==> T = 2*0.9/V0 = 1.8/30 = 0.06 H.
Then, a = -V0/T = -30/0.06 = -500 km/h².
So, stop time is T = 0.06 H and retarsation is a = -500 km/h².

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