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Answer:
Approximately [tex]0.914[/tex] miles.
Explanation:
Since acceleration is a constant value, the average velocity of the vehicle during this period of [tex]t = 1.2\; \text{minute}[/tex] would be the same as the average of its velocity before and after the acceleration:
[tex]\displaystyle \bar{v} = \frac{u + v}{2}[/tex],
Where:
To find the displacement of the vehicle [tex]x[/tex], multiply the duration of the period by the average velocity of the vehicle during this period:
[tex]\begin{aligned} x = \bar{v}\, t = \left(\frac{u + v}{2}\right)\, t\end{aligned}[/tex].
Make sure that the units are consistent. Specifically, while the unit for time is minutes in the given value of the duration, the unit for time in the given values of velocity is hours. Apply unit conversion and ensure that the duration of the motion is measured in hours:
[tex]\displaystyle t = 1.2\; \text{minute} \times \frac{1\; \text{hour}}{60\; \text{minute}} = 0.02\; \text{hour}[/tex].
Substitute the values into the expression for displacement:
[tex]\begin{aligned} x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{33.4\; \text{mph} + 58.0\; \text{mph}}{2} \times 0.02\; \text{hour} \\ &= 0.914\; \text{mile} \end{aligned}[/tex].