Question:
A car covers a distance at speed of \(60\text{ km h}^{-1}\). It returns and comes back to the original point moving at a speed of \(V\). If the average speed for the round trip is \(48\text{ km h}^{-1}\), then the magnitude of \(V\) is
A car covers a distance at speed of \(60\text{ km h}^{-1}\). It returns and comes back to the original point moving at a speed of \(V\). If the average speed for the round trip is \(48\text{ km h}^{-1}\), then the magnitude of \(V\) is
Show Hint
For equal distances, average speed is not the arithmetic mean. Use
\[
\text{Average speed}=\frac{2uv}{u+v}
\]
which is the harmonic mean of the two speeds.
Updated On: Jun 15, 2026
- \(40\text{ km h}^{-1}\)
- \(36\text{ km h}^{-1}\)
- \(44\text{ km h}^{-1}\)
- \(32\text{ km h}^{-1}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use the formula for average speed.
For equal distances covered with speeds \(u\) and \(v\), the average speed is \[ \text{Average speed}=\frac{2uv}{u+v} \] Here, \[ u=60\text{ km h}^{-1} \] and \[ v=V \] Given average speed, \[ 48=\frac{2(60)V}{60+V} \]
Step 2: Simplify the equation.
\[ 48(60+V)=120V \] \[ 2880+48V=120V \] \[ 2880=72V \] \[ V=\frac{2880}{72} \] \[ V=40 \]
Step 3: Final Answer.
Therefore, \[ \boxed{40\text{ km h}^{-1}} \]
For equal distances covered with speeds \(u\) and \(v\), the average speed is \[ \text{Average speed}=\frac{2uv}{u+v} \] Here, \[ u=60\text{ km h}^{-1} \] and \[ v=V \] Given average speed, \[ 48=\frac{2(60)V}{60+V} \]
Step 2: Simplify the equation.
\[ 48(60+V)=120V \] \[ 2880+48V=120V \] \[ 2880=72V \] \[ V=\frac{2880}{72} \] \[ V=40 \]
Step 3: Final Answer.
Therefore, \[ \boxed{40\text{ km h}^{-1}} \]
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