Answer to Question #169753 in Physics for Festus

Question

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/hr. Finding the market closed, he instantly turns and walk back home with a speed of 7.5km/her. The average speed of the man over the interval of time 0 to 40 mins is equal to____

Answer 1

The average speed can be found as follows:

$$v_{avg}=\dfrac{d_{tot}}{t_{tot}},$$ $$v_{avg}=\dfrac{d_1+d_2}{t_1+t_2}.$$

The distance traveled to the market:

$$d_1=2.5\ km.$$

Let's find time taken to go to the market:

$$t_1=\dfrac{d_1}{v_1}=\dfrac{2.5\ km}{5\ \dfrac{km}{h}}=\dfrac{1}{2}\ h=30\ min.$$

Let's find time taken to return back:

$$t_2=\dfrac{d_2}{v_2}=\dfrac{2.5\ km}{7.5\ \dfrac{km}{h}}=\dfrac{1}{3}\ h=20\ min.$$

But, in fact, we need his average speed in the interval of time 0 to 40 minutes, so $t_2=10\ min$.

Let's find the distance traveled in 10 minutes while the man returning back:

$$d_2=v_2t_2=7.5\ \dfrac{km}{h}\cdot10\ min\cdot\dfrac{1\ h}{60\ min}=1.25\ km.$$

Finally, we can find the average speed of the man over the interval of time 0 to 40 minutes:

$$v_{avg}=\dfrac{2.5\ km+1.25\ km}{40\ min\cdot\dfrac{1\ h}{60\ min}}=5.625\ \dfrac{km}{h}.$$