Average speed in a circular trajectory

First, we must know the length of the circumference.

$$L=2\mathrm{\pi r}=2\mathrm{\pi }80=160\mathrm{\pi }$$ 

1.- The distance traveled ∆s will be one fourth of the circumference length. Remember that the distance traveled is a scalar, unlike the displacement which is a vector. We simply divide the length by four to obtain the distance traveled.

$$∆s=\frac{L}{4}=40\mathrm{\pi }\text{m}$$ 

Finally, by applying the formula of the average speed we get

$$V_{\mathrm{avg}}=\frac{∆s}{∆t}=\frac{40\mathrm{\pi }}{13}=\boxed{9.66 \text{m/s}}$$

2.- In the second question we are asked for the average speed for a distance that is double than the distance in question one, the time taken to travel that distance is also double so, the average speed should be the same. Doing the calculation.

In this case, the distance traveled ∆s, will be half the length of the circumference. Simply, we divide the length by two to obtain the distance traveled.

$$∆s=\frac{L}{2}=80\mathrm{\pi }\text{m}$$ 

Finally, applying the average speed formula we get:

$$V_{\mathrm{avg}}=\frac{∆s}{∆t}=\frac{80\mathrm{\pi }}{26}=\boxed{9.66 \text{m/s}}$$

3.- Finally, we are asked the average speed for a distance which is the same as in question two (double than in question 1), in half the time of question 2 (the same as in question 1). From this we can infer that the average speed will be double than that in question 2.

In this case, distance traveled ∆s, shall be equal to the one in question 2.

Applying the average speed formula, we get:

$$V_{\mathrm{avg}}=\frac{∆s}{∆t}=\frac{80\mathrm{\pi }}{13}=\boxed{19.32 \text{m/s}}$$