A particle moves on a straight line in such way that it covers $1 s t$ half distance with speed $3 m / s$ and next half distance in $2$ equal time intervals with speeds $4.5 m / s$ and $7.5 m / s$ respectively. Find average speed of the particle.

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Solution

Suppose total distance is $2 x$
then first half is $2 x$ traveled with $3 m / s$ so time taken $t_{1} = \frac{2 x}{3}$
after it suppose it take $2 t$ time then
its half i.e. $t$ with $4.5 m / s$ so distance $s_{1} = 4.5 t$
another half i.e. $t$ with $7.5 m / s$ so distance $s_{2} = 7.5 t$
this distance $s_{1} + s_{2} = 2 x$ or $4.5 t + 7.5 t = 12 t = 2 x$ or $x = 6 t$

Average speed is $v = \frac{d i s t a n c e}{t i m e} = \frac{4 x}{t_{1} + 2 t} = \frac{4 x}{\frac{2 x}{3} + 2 t}$ putting $x = 6 t$ we get $v = 4 m / s$

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