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Sagot :
Let's use the variable x to represent the average speed and d to represent the total distance.
Then, let's calculate the time needed for each part:
The first third has a speed of 2 mph, so the time is:
[tex]\begin{gathered} \text{distance}=\text{speed}\cdot\text{time} \\ \frac{d}{3}=2\cdot t \\ t=\frac{d}{6} \end{gathered}[/tex]For the second third, we have a speed of 3 mph, so:
[tex]\begin{gathered} \frac{d}{3}=3\cdot t \\ t=\frac{d}{9} \end{gathered}[/tex]The last third has a speed of x, so:
[tex]\begin{gathered} \frac{d}{3}=x\cdot t \\ t=\frac{d}{3x} \end{gathered}[/tex]Then, the average speed is the total distance over the total time, so:
[tex]\begin{gathered} x=\frac{d}{t_{\text{total}}} \\ x=\frac{d}{\frac{d}{6}+\frac{d}{9}+\frac{d}{3x}} \\ x=\frac{1}{\frac{1}{6}+\frac{1}{9}+\frac{1}{3x}} \\ x=\frac{1}{\frac{3x+2x+6}{18x}} \\ x=\frac{18x}{5x+6} \\ 1=\frac{18}{5x+6} \\ 5x+6=18 \\ 5x=12 \\ x=2.4 \end{gathered}[/tex]The average speed of the trip is 2.4 m/s.
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