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A car initially at rest undergoes uniform acceleration for 6.32 seconds and covers a distance of 120 meters. What is the approximate acceleration of the car?

Verified Answer :

AbhiGhost

Using kinematic equation s=ut + 1/2 at^2(u = initial velocity=0, s=120m, t= 6.32s), 120 = 0(t) + 1/2 a(6.32)^2. a = 120x2/(6.32)^2 = 6m/s^2.

Answer : The approximate acceleration of the car is, \(6.008m/s^2\)

Solution : Given,

Initial velocity of the car = 0 m/s

Distance covered = 120 meter

Time taken = 6.32 second

Using second law of motion,

\(s=ut+\frac{1}{2}at^2\)

where,

s = distance covered

u = initial velocity

t = time taken

a = acceleration

Now put all the given values in the above equation, we get

\((120m)=(0m/s)\times (6.32s)+\frac{1}{2}\times a\times (6.32s)^2\)

\(a=6.008m/s^2\)

Therefore, the approximate acceleration of the car is, \(6.008m/s^2\)

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