A cyclist travels a distance of $1 k m$ in the first hour, $0.5 k m$ in the second hour and $0.3 k m$ in the third hour. What is the average speed in $(1) k m / h r (2) m / s e c ?$ ?

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Solution

Step-1 Finding the average speed in $k m / h r$
Distance in one hour $= 1 k m$
Distance in second hour $= 0.5 k m$
Distance in third hour $= 0.3 k m$
We know that,
$A v e r a g e s p e e d = \frac{T o t a l d i s \tan c e}{T o t a l t i m e}$
Average speed in $k m / h r$
$\begin{array}{rcl} ν & = & \frac{D}{T} \\ = & \frac{(1 + 0.5 + 0.3)}{3} k m / h r \\ = & 0.6 k m / h r \end{array}$
Step-2 Finding the average speed in $m / s e c$
Average speed in $m / s e c$
$\begin{array}{rcl} ν & = & \frac{D}{T} \\ = & \frac{(1 + 0.5 + 0.3)}{3} k m / h r \\ = & 0.6 k m / h r \\ = & 0.6 \times \frac{5}{18} m / s e c (1 k m / h r = \frac{5}{18} m / s e c) \\ = & 0.167 m / s e c \end{array}$
Therefore, Average speed is $0.6 k m / h r$ and $0.167 m / s e c$ .

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