average speed.

• Sep 17th 2007, 01:45 PM
hana_102
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
• Sep 17th 2007, 03:23 PM
Soroban
Hello, hana_102!

This is a classic Trick Question . . .

Quote:

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?

$\text{Average Speed} \:= \:\frac{\text{Total distance}}{\text{Total Time}}$

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

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

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

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

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

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

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

• Sep 18th 2007, 07:25 AM
hana_102
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
• Sep 18th 2007, 08:28 AM
Soroban
Hello, hana_102!

Don't kick yourself too hard . . .

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

• Sep 18th 2007, 12:01 PM
hana_102
Thank you. That seems so simple now that I look at it all worked out!!!!!!!! :eek: