# Question 4aa5a

Sep 4, 2015

48 km/hr

#### Explanation:

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 $120 \times 2 = 240$ km
color(white)("XXXX")3 " hours" + 2 " hours" = 5 " hours"

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

Sep 4, 2015

The answer is (b) $\text{48 km/h}$.

#### Explanation:

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 $\text{40 km/h}$, and ${t}_{2}$ hours to travel the same distance at $\text{60 km/h}$.

This means that you can write

$d = 40 \cdot {t}_{1} \text{ }$ and $\text{ } d = 60 \cdot {t}_{2}$

This is equivalent to

$40 \cdot {t}_{1} = 60 \cdot {t}_{2} \implies {t}_{1} = \frac{3}{2} \cdot {t}_{2}$

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

${d}_{\text{total}} = 2 \cdot d$

The total time will be

${t}_{\text{total}} = {t}_{1} + {t}_{2}$

${t}_{\text{total}} = \frac{3}{2} \cdot {t}_{2} + {t}_{2} = \frac{5}{2} \cdot {t}_{2}$

The average speed will thus be

$\overline{v} = \frac{2 d}{\frac{5}{2} \cdot {t}_{2}} = \frac{4}{5} \cdot \frac{d}{t} _ 2$

But $\frac{d}{t} _ 2$ is equal to the speed of his return trip, $\text{60 km/h}$. This means that you have

$\overline{v} = \frac{4}{5} \cdot 60 = \textcolor{g r e e n}{\text{48 km/h}}$