Question

Going from point A to point B, the cheetah traveled at an average rate of 70 mph. Returning to point A, the cheetah traveled at an average rate of 40 mph. Can we say that this cheetah’s average rate was 55 mph? The average rate for the trip is equal to the total distance traveled divided by the total time traveled. The following equations represent the distance traveled on each leg of the trip. First leg of trip: d=r_1 t_1 Second leg of trip: d=r_2 t_2 Write an equation for the average rate for the trip. Remember, the cheetah runs from point A to point B and back to point A.

Answer 1

Answer:

No.

$\text{Average rate} = \dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{r_{1}t_{1} + r_{2}t_{2}}{t_{1} + t_{2}}\\\\= \text{50.91 mi/h}$

Step-by-step explanation:

First leg:              d = r₁t₁

Second leg:        d = r₂t₂

                         r₁t₁ = r₂t₂

Total distance: 2d = r₁t₁ + r₂t₂

Total time:           t = t₁ + t₂

1. Equation for average rate

$\text{Average rate} = \dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{r_{1}t_{1} + r_{2}t_{2}}{t_{1} + t_{2}}$

2. Average rate

Since r₁t₁ = r₂t₂,

$t_{1} = \dfrac{r_{2}t_{2}}{r_{1}} = \frac{40}{70}t_{2} = \frac{4}{7}t_{2}\\\\\text{Average rate} = \dfrac{2r_{2}t_{2}}{\frac{4}{7}t_{2} + t_{2}} = \dfrac{2 \times 40t_{2}}{\frac{4t_{2}+ 7t_{2}}{7}}= 80t_{2} \times\frac{7}{11t_{2}} = \dfrac{560}{11}\\\\= \textbf{50.91 mi/h}$

We cannot say truthfully that the average rate is 55 mi/h.