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
[tex]d \to[/tex] distance
[tex]r \to[/tex] average rate
[tex]t \to[/tex] time
We have:
[tex]r_1 =70[/tex]
[tex]r_2 = 40[/tex]
[tex]d = r_1t_1 = r_2 t_2[/tex]
The average rate of the trip is:
[tex]r = \frac{2d}{t}[/tex]
Where
[tex]2d \to[/tex] the total distance from A to B and back to A
[tex]2d = r_1t_1 + r_2t_2[/tex]
and
[tex]t \to[/tex] total time spent from A to B and back to A
[tex]t =t_1 + t_2[/tex]
[tex]r = \frac{2d}{t}[/tex] becomes
[tex]r = \frac{r_1t_1 + r_2t_2}{t_1 + t_2}[/tex]
Recall that:
[tex]d = r_1t_1 = r_2 t_2[/tex]
Make [tex]t_2[/tex] the subject
[tex]t_2 = \frac{r_1t_1}{r_2}[/tex]
Substitute [tex]t_2 = \frac{r_1t_1}{r_2}[/tex] in [tex]r = \frac{r_1t_1 + r_2t_2}{t_1 + t_2}[/tex]
[tex]r = \frac{r_1t_1 + r_2\times \frac{r_1t_1}{r_2}}{t_1 + \frac{r_1t_1}{r_2}}[/tex]
Factor out t1
[tex]r = \frac{t_1(r_1 + r_2\times \frac{r_1}{r_2})}{t_1(1 + \frac{r_1}{r_2})}[/tex]
[tex]r = \frac{r_1 + r_2\times \frac{r_1}{r_2}}{1 + \frac{r_1}{r_2}}[/tex]
[tex]r = \frac{r_1 + r_1}{1 + \frac{r_1}{r_2}}[/tex]
Substitute values for r1 and r2
[tex]r = \frac{70 + 70}{1 + \frac{70}{40}}[/tex]
[tex]r = \frac{140}{1 + 1.75}[/tex]
[tex]r = \frac{140}{2.75}[/tex]
[tex]r = 50.91[/tex]
Hence
- The equation of the average rate is: [tex]r = \frac{r_1t_1 + r_2t_2}{t_1 + t_2}[/tex].
- The average rate is 50.91mph, not 55mph
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