# 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.

Expert's answer

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.

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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