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A motor cyclist travels a certain distance with a uniform speed of $30\, m/s$ and immediately turns back and returns to starting point with a uniform speed $20\, m/s$. Then the average speed of the motor cycle is ..........$m/s$ :-
$12$
$50$
$24$
$25$
Solution
Let distance between $\mathrm{A}$ and $\mathrm{B}$ is $\mathrm{D}$ Then, time taken by motorcyclist with uniform speed $30$ $\mathrm{m} / \mathrm{s}, \mathrm{T}_{1}$ $=$ distance/speed $=\mathrm{D} / 30 \mathrm{hrs}$
Again, they return with uniform speed $20 \mathrm{m},$ so time taken by motorcyclist, $\mathrm{T}_{2}=\mathrm{D} / 20 \mathrm{hrs}$
Now, average speed of motorcyclist $=$ total distance/total time taken $=(D+D) /(D / 30+D / 20) \mathrm{m} / \mathrm{s}$
$=2 \times 30 \times 20 /(30+20) \mathrm{m} / \mathrm{s}$
$=24 \mathrm{m} / \mathrm{s}$
