Question 49 of 308

A horse rider covers half the distance with $5m{s}^{-1}$ speed. The remaining part of the distance was travelled with speed $10m{s}^{-1}$ for half the time and with speed $15m{s}^{-1}$ for other half of the time. The mean speed of the rider averaged over the whole time of motion is $\dfrac{x}{7}m{s}^{-1}$. The value of $x$ is ______.

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Answer: 50

Let total distance travelled by horse rider be $2x$.

Now, the time taken to cover first half distance is ${t}_{AB}=\dfrac{x}{5}$
Second half distance $x$ between $BC$ is covered in two parts, $x={d}_{1}+{d}_{2}$
here ${d}_{1}$ is the distance travelled with $10m{s}^{-1}$ and ${d}_{2}$ is the distance travelled with $15m{s}^{-1}$ in equal time intervals $\dfrac{t}{2}$ each.
Then, ${d}_{1}=\dfrac{10t}{2},{d}_{2}=\dfrac{15t}{2}$
Second half distance $x=\dfrac{t}{2}(10+15)=\dfrac{25t}{2}$
Now, the mean speed of rider is $v=\dfrac{2x}{\dfrac{x}{5}+\dfrac{2x}{25}}=\dfrac{2\times 25}{5+2}=\dfrac{50}{7}m{s}^{-1}$
Thus, the value of $x=50$.