A particle moving along a straight line covers $$60\%$$ of the distance at $$v$$. If the speed averaged for the entire distance is $$\frac{5v}{4}$$, then the speed with which the particle travelled for the remaining distance is

Solution

Let the total distance be $$d$$ and the speed on the last $$40\%$$ be $$u$$. Average speed $$= \frac{d}{\frac{0.6d}{v} + \frac{0.4d}{u}} = \frac{5v}{4}$$, so $$\frac{0.6}{v} + \frac{0.4}{u} = \frac{0.8}{v}$$. This gives $$\frac{0.4}{u} = \frac{0.2}{v}$$ and $$u = 2v$$.

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