A car undergoes a constant acceleration from speed of 33.4 miles per hour to speed of 58.0 miles per hour while moving for a time interval of 1.2 minutes along a straight line. What distance in miles does the car travel during this time?

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:

[tex]\bar{v}[/tex] is average velocity in this period,
[tex]u[/tex] is the velocity before the acceleration, and
[tex]v[/tex] is the velocity after the acceleration.

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].