1. ## average speed.

“if a cyclist travels at an average speed of 3m/s-1 and turns home at an average speed of 9m/s-1 than what is the average speed of the whole journey?”

using the equation speed= distance/time

2. Hello, hana_102!

This is a classic Trick Question . . .

If a cyclist travels at an average speed of 3m/s
and turns home at an average speed of 9m/s,
what is the average speed of the whole journey?
$\displaystyle \text{Average Speed} \:= \:\frac{\text{Total distance}}{\text{Total Time}}$

We will use: .$\displaystyle Time\;=\;\frac{Distance}{Speed}$

Suppose the distance (one way) is $\displaystyle d$ meters.

He cycled $\displaystyle d$ meters at 3 m/s.
. . This took him: .$\displaystyle \frac{d}{3}$ seconds.

He cycled back $\displaystyle d$ meters at 9 m/s.
. . This took him: .$\displaystyle \frac{d}{9}$ seconds.

The entire trip took: .$\displaystyle \frac{d}{3} + \frac{d}{9} \:=\:\frac{4}{9}d$ seconds (Total time).

The total distance is: .$\displaystyle 2d$ meters.

Therefore, his average speed was: .$\displaystyle \frac{2d}{\frac{4}{9}d} \;=\;4.5\text{ m/s}$

3. Thanks for explaining, I understand how you got your answer, but what I don’t get is how you got 4/9 ?
Sorry if this seems silly. But how does d/3 + d/9= 4/9d?

Thank you

4. Hello, hana_102!

Don't kick yourself too hard . . .

. . $\displaystyle \frac{1}{3} + \frac{1}{9} \;=\;\frac{3}{9} + \frac{1}{9} \;=\;\frac{4}{9}$

5. Thank you. That seems so simple now that I look at it all worked out!!!!!!!!