Answer:
No.
[tex]\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}[/tex]
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
[tex]\text{Average rate} = \dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{r_{1}t_{1} + r_{2}t_{2}}{t_{1} + t_{2}}[/tex]
2. Average rate
Since r₁t₁ = r₂t₂,
[tex]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}[/tex]
We cannot say truthfully that the average rate is 55 mi/h.
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
Given that:
[tex]d = r_1t_1 = r_2t_2[/tex]
So, we have:
[tex]r_1 = 70[/tex]
[tex]r_2 = 40[/tex]
The average rate (r) is calculated as follows:
[tex]r = \frac{Total\ Distance (D)}{Total\ Time (T)}[/tex]
Where:
[tex]D=d + d[/tex]
[tex]D = r_1t_1 + r_2t_2[/tex]
and
[tex]T =t_1 + t_2[/tex]
So, the average rate is:
[tex]r =\frac{r_1t_1 + r_2t_2}{t_1 + t_2}[/tex]
Recall that:
[tex]D = r_1t_1 + r_2t_2[/tex]
[tex]r_1t_1 = r_2t_2[/tex]
Make [tex]t_2[/tex] the subject
[tex]t_2 = \frac{r_1t_1}{r_2}[/tex]
Substitute values for [tex]r_1[/tex] and [tex]r_2[/tex]
[tex]t_2 = \frac{70t_1}{40}[/tex]
So, we have:
[tex]r =\frac{r_1t_1 + r_2t_2}{t_1 + t_2}[/tex]
[tex]r = \frac{70t_1 + 40 \times \frac{70t_1}{40}}{t_1 + \frac{70t_1}{40}}[/tex]
[tex]r = \frac{70t_1 + 70t_1}{t_1 + \frac{70t_1}{40}}[/tex]
[tex]r = \frac{140t_1}{t_1 + \frac{70t_1}{40}}[/tex]
Factor out t1
[tex]r= \frac{140t_1}{t_1(1 + \frac{70}{40})}[/tex]
[tex]r = \frac{140}{(1 + \frac{70}{40})}[/tex]
[tex]r = \frac{140}{\frac{40+70}{40}}[/tex]
[tex]r= \frac{140}{\frac{110}{40}}[/tex]
Rewrite as:
[tex]r = \frac{140 \times 40}{110}[/tex]
[tex]r = 50.91mph[/tex]
Hence, the cheetah's average rate is 50.91mph, not 55mph
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