The average speed of Joshua during that time is 2500 m/h.
Explanation:
It is given that Joshua started cycling at 5:15 pm. By 8:09 pm he has covered a distance of 7250 m.
The total time taken by Joshua from 5:15 pm to 8:09 pm is
[tex]2 { h } 54 \text { min }=2 \mathrm{h}+\frac{54}{60} {h}[/tex]
Dividing we get,
[tex]2 { h } 54 \text { min }=2 \mathrm{h}+0.9 {h}[/tex]
Adding, we have,
[tex]2 { h } 54 \text { min }=2.9 {h}[/tex]
Thus, the total time taken by Joshua is [tex]2.9 {h}[/tex]
To determine the average speed we use the formula,
[tex]speed=\frac{distance}{time}[/tex]
where [tex]distance=7250[/tex] and [tex]time=2.9[/tex]
Hence, substituting the values we have,
[tex]speed=\frac{7250}{2.9}[/tex]
Dividing, we get,
[tex]speed=2500[/tex]
Thus, the average speed of Joshua during that time is 2500 m/h.
