“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

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- Sep 17th 2007, 01:45 PMhana_102average 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 PMSoroban
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?

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: .$\displaystyle \frac{2d}{\frac{4}{9}d} \;=\;4.5\text{ m/s}$

- Sep 18th 2007, 07:25 AMhana_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 AMSoroban
Hello, hana_102!

Don't kick yourself too hard . . .

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

- Sep 18th 2007, 12:01 PMhana_102
Thank you. That seems so simple now that I look at it all worked out!!!!!!!! :eek: