Question

Step 3: Calculating the average rate Now it’s time to tackle the original 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 a). 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=r1t1 Second leg of trip: d=r2t2 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. The following equation represents the average rate for the trip. Is this equation equivalent to the one you wrote above? Explain why or why not. average rate= 2d/d/r1+d/r2 The following steps show how the equation for the average rate can be transformed so that it is written in terms of only the rates for each leg of the trip. Write an algebraic justification for each step. Think about the number operations and properties that you know. average rate =2d/d/r1+d/r2 average rate= 2d/r2d/r1r2+r1d/r1r2 = 2d /r1d+r2d//r1r2 average rate= 2d/r1d+r2d/r1r2 = 2d/1 times r1r2/r1d+r2d = 2dr1r2/r1d+r2d average rate= 2dr1r2/r1d+r2d = 2dr1r2/d(r1+r2) = 2r1r2/r1+r2 Using the equation for average rate above, determine the cheetah’s average rate for the entire trip.

Answer 1

The average rate of a body is the total distance travelled divided by the total time.

The cheetah runs at an average rate of 50.91mph, not 55mph

Let

$d \to$ distance

$r \to$ average rate

$t \to$ time

We have:

$r_1 =70$

$r_2 = 40$

$d = r_1t_1 = r_2 t_2$

The average rate of the trip is:

$r = \frac{2d}{t}$

Where

$2d \to$ the total distance from A to B and back to A

$2d = r_1t_1 + r_2t_2$

and

$t \to$ total time spent from A to B and back to A

$t =t_1 + t_2$

$r = \frac{2d}{t}$ becomes

$r = \frac{r_1t_1 + r_2t_2}{t_1 + t_2}$

Recall that:

$d = r_1t_1 = r_2 t_2$

Make $t_2$ the subject

$t_2 = \frac{r_1t_1}{r_2}$

Substitute $t_2 = \frac{r_1t_1}{r_2}$ in $r = \frac{r_1t_1 + r_2t_2}{t_1 + t_2}$

$r = \frac{r_1t_1 + r_2\times \frac{r_1t_1}{r_2}}{t_1 + \frac{r_1t_1}{r_2}}$

Factor out t1

$r = \frac{t_1(r_1 + r_2\times \frac{r_1}{r_2})}{t_1(1 + \frac{r_1}{r_2})}$

$r = \frac{r_1 + r_2\times \frac{r_1}{r_2}}{1 + \frac{r_1}{r_2}}$

$r = \frac{r_1 + r_1}{1 + \frac{r_1}{r_2}}$

Substitute values for r1 and r2

$r = \frac{70 + 70}{1 + \frac{70}{40}}$

$r = \frac{140}{1 + 1.75}$

$r = \frac{140}{2.75}$

$r = 50.91$

Hence

• The equation of the average rate is: $r = \frac{r_1t_1 + r_2t_2}{t_1 + t_2}$.
• The average rate is 50.91mph, not 55mph

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