Question

A person walks first at a constant speed of 4.50 m/s along a straight line from point circle a to point circle b and then back along the line from circle b to circle a at a constant speed of 2.95 m/s. (a) What is her average speed over the entire trip?

Answer 1

OK, so you want to find the average speed. In that case, let's start looking at the equation to find average speed.\[\text{avg. speed} = \frac{\text{total distance travelled}}{\text{total time}}\]
Right? Now let's see if we can find seperate equations for distance and time.

Answer 2

Now, we know that she walks back and forth the same way, so we can argue that the distance travelled one way, is exactly as far as the distance travelled the other way. But because she walks at different speeds, the time will also be different, so let make two equations for the time travelled.
\[t_1=\frac{d}{v_1} \text{ and } t_2=\frac{d}{v_2}\]
This makes sense, right?

Answer 3

So, let's put all that we know in the original equation.
\[\text{avg.speed}=\frac{2 \times d}{t_1+t_2}\]
This means, total distance travelled is twice the distance of one way. And the total time is t1 plus t2, right?

Answer 4

So, lets put our equations for time into t1 and t2.
\[\text{avg. speed}=\frac{2 \times d}{\frac{d}{v_1}+\frac{d}{v2}}\]
The d is both in the numerator and the denominator, so let's cancel it out.
\[\text{avg. speed}=\frac{2}{\frac{1}{v_1}+\frac{1}{v2}}\]
Even though you could actually find your answer now, I prefer to clean up the equation, so lets remove the fractions from the denominator.
\[\text{avg. speed}=\frac{2}{\frac{1}{v_1}+\frac{1}{v2}} \times \frac{v_1v_2}{v_1v_2}=\frac{2 \times v_1v_2}{v_2+v_1}\]
So, now you have a simple equation for avg. speed:
\[\text{avg. speed}=\frac{2 \times v_1v_2}{v_2+v_1}\]
I hope you understood how to find this. Otherwise, just ask! :)