Question #4aa5a

Pick some distance the motorcyclist might have traveled (it doesn't matter what distance but for ease of calculation I chose 120 km).

Time spent traveling 120 km at 40 km/hr
`color(white)("XXXX")120" km" div 40 " km/hr" = 3 " hours"`

Time spent traveling 120 km at 60 km/hr
`color(white)("XXXX")120" km" div 60 " km/hr" = 2 " hours"`

Total time spent traveling `120 xx 2 = 240` km
`color(white)("XXXX")3 " hours" + 2 " hours" = 5 " hours"`

Average speed
`color(white)("XXXX")240" km" div 5" hours" = 48 " km/hr"`

I want to provide an alternative approach to figuring out the average sspeed of the motorcyclist.

So, you know that the motorcyclist travels the same distance, let's say `d`, for both his trips.

Let's assume that it takes him `t_1` hours to travel the distance at `"40 km/h"`, and `t_2` hours to travel the same distance at `"60 km/h"`.

This means that you can write

`d = 40 * t_1" "` and `" "d = 60 * t_2`

This is equivalent to

`40 * t_1 = 60 * t_2 implies t_1 = 3/2 * t_2`

The average speed of the motorcyclist can be thought of as the total distance he covered divided by the total time needed.

Since he goes from `A` to `B` for one trip, and from `B` to `A` for the second, the total distance covered will be

`d_"total" = 2 * d`

The total time will be

`t_"total" = t_1 + t_2`

`t_"total" = 3/2 * t_2 + t_2 = 5/2 * t_2`

The average speed will thus be

`bar(v) = (2d)/(5/2 * t_2) = 4/5 * d/t_2`

But `d/t_2` is equal to the speed of his return trip, `"60 km/h"`. This means that you have

`bar(v) = 4/5 * 60 = color(green)("48 km/h")`