A man walks $7 k m$ in $2 h o u r s$ and $2 k m$ in $1 h o u r$ in the same direction. What is man's average velocity for the whole journey?

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Solution

Step 1: Given
Distance covered in 2 hours: $D_{1} = 7 k m$
Distance covered in 1 hour: $D_{2} = 2 k m$
Time taken to cover D₁: $T_{1} = 2 h o u r s$
Time taken to cover D₂: $T_{2} = 1 h o u r$
Step 2: Formula Used
$A v e r a g e V e l o c i t y = \frac{T o t a l D i s p l a c e m e n t}{T o t a l T i m e}$
Step 3: Find the average velocity
Calculate the total distance by adding both the distances.
$T o t a l D i s \tan c e = D_{1} + D_{2} = 7 k m + 2 k m = 9 k m$
Calculate the total time by adding both the times.
$T o t a l T i m e = T_{1} + T_{2} = 2 h r s + 1 h r = 3 h r s$
Calculate the average velocity using the formula. Since the motion is in a straight line, $D i s p l a c e m e n t = D i s \tan c e$ .
$A v e r a g e V e l o c i t y = \frac{T o t a l D i s p l a c e m e n t}{T o t a l T i m e} = \frac{9 k m}{3 h r s} = 3 k m / h r s$
Hence, the average velocity is 3 km/hrs.

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