The average of Jamal on the trip to school and back home is 4.6 milles per hour.
Let consider that Jamal moves at uniform speed, in miles per hour. The average speed is equal to the following formula:
[tex]v = \frac{x_{F}+x_{B}}{t_{F}+t_{B}}[/tex] (1)
Where:
- [tex]x_{F}[/tex] - Forward distance, in miles.
- [tex]x_{B}[/tex] - Backward distance, in miles.
- [tex]t_{F}[/tex] - Forward time, in hours.
- [tex]t_{B}[/tex] - Backward time, in hours.
Forward and backward times are described by the following formulas:
[tex]t_{F} = \frac{x_{F}}{v_{F}}[/tex] (2)
[tex]t_{B} = \frac{x_{B}}{v_{B}}[/tex] (3)
If we know that [tex]x_{F} = x_{B} = 4\,mi[/tex], [tex]v_{F} = 10\,\frac{mi}{h}[/tex] and [tex]v_{B} = 3\,\frac{mi}{h}[/tex], then the average speed of Jamal is:
[tex]t_{F} = \frac{x_{F}}{v_{F}}[/tex]
[tex]t_{F} = \frac{4\,mi}{10\,\frac{mi}{h} }[/tex]
[tex]t_{F} = 0.4\,h[/tex]
[tex]t_{B} = \frac{x_{B}}{v_{B}}[/tex]
[tex]t_{B} = \frac{4\,mi}{3\,\frac{mi}{h} }[/tex]
[tex]t_{B} = 1.333\,h[/tex]
[tex]v = \frac{8\,mi}{0.4\,h + 1.333\,h}[/tex]
[tex]v = 4.616\,\frac{mi}{h}[/tex]
The average of Jamal on the trip to school and back home is 4.6 milles per hour.
We kindly invite to check this question on average speed: https://brainly.com/question/24928084
