Question #126340

A car traveling at 50 metre per seconds from rest covers a distance of 10km calculate the acceleration

Expert's answer

Assuming the constant acceleration, the covered distance is given by:


d=at22d = \dfrac{at^2}{2}

where d=10×103md = 10\times10^3m is the distance, aa is the acceleration and tt is the time.

On the other hand, by definition, the acceleration is:


a=v0t=vta = \dfrac{v - 0}{t} = \dfrac{v }{t}

where v=50m/sv = 50m/s is the final velocity and stands for the intitial velocity (start from rest).

Substituting the expression for aa into the first equation and expressing tt , get:


d=vt22t=vt2t=2dvd = \dfrac{vt^2}{2t} = \dfrac{vt}{2}\\ t = \dfrac{2d}{v}

Substituting the last equation into the expression for aa, obtain:


a=v22d=502210103=0.125m/s2a = \dfrac{v^2}{2d} = \dfrac{50^2}{2\cdot 10\cdot 10^3} = 0.125 m/s^2

Answer. 0.125 m/s^2.


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