“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
Hello, hana_102!
This is a classic Trick Question . . .
$\displaystyle \text{Average Speed} \:= \:\frac{\text{Total distance}}{\text{Total Time}} $If a cyclist travels at an average speed of 3m/s
and turns home at an average speed of 9m/s,
what is the average speed of the whole journey?
We will use: .$\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: .$\displaystyle \frac{2d}{\frac{4}{9}d} \;=\;4.5\text{ m/s}$