Question

A cyclist travels a distance of 1 km in the first hour, 0.5 km in the second hour and 0.3 km in the third hour. Find the average speed of the cyclist in​
(i)$km{h}^{-1}$​
(ii) $m{s}^{-1}$. ​

Answer 1

Given,
Total distance travelled = (1+0.5+0.3) km​ = 1.8 km​

Time taken taken = (1+1+1) h = 3 h​
To find, Average speed​
(i) Average speed (in $km{h}^{-1})$​
$\text{Average speed}=\frac{\text{Total distance covered}}{\text{Total time taken}}$

$=\frac{1.8km}{3h}$
The average speed of cyclist is $0.6km{h}^{-1}$.​

(ii) Average speed (in $m{s}^{-1})$​
Total distance travelled $=1.8km=1800m$
Time taken taken $=3h=10800s$
$\text{Average speed}=\frac{\text{Total distance covered}}{\text{Total time taken}}$
$=\frac{1800m}{10800s}$
The average speed of cyclist is $0.167km{h}^{-1}$.​