Time is given as:

$\overline{){\mathbf{t}}{\mathbf{=}}\frac{\mathbf{d}}{\mathbf{v}}}$, where d is distance and v is speed.

**(a)**

The time the car takes in the first part of the motion is given by:

${{\mathbf{t}}}_{{\mathbf{1}}}{\mathbf{=}}\frac{\mathbf{d}}{{\mathbf{v}}_{\mathbf{1}}}$

Let the motion in the second part be described by time, t_{2,} and constant speed v_{2}.

The time the car takes in this part is given by:

${\mathit{t}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{d}}{{\mathbf{v}}_{\mathbf{2}}}$

The average velocity in the whole motion is:

$\begin{array}{rcl}{\mathbf{v}}_{\mathbf{a}\mathbf{v}\mathbf{g}}& \mathbf{=}& \frac{\mathbf{d}\mathbf{+}\mathbf{d}}{{\mathbf{t}}_{\mathbf{1}}\mathbf{+}{\mathbf{t}}_{\mathbf{2}}}\\ & \mathbf{=}& \frac{\mathbf{2}\mathbf{d}}{{\mathbf{t}}_{\mathbf{1}}\mathbf{+}{\mathbf{t}}_{\mathbf{2}}}\end{array}$

A car travels along a straight line at a constant speed of 60.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 30.0 mi/h.

(a) What is the constant speed with which the car moved during the second distance d?

(b) What If? Suppose the second distance d were traveled in the opposite direction you forgot something and had to return home at the same constant speed as found in part (a). What is the average velocity for this trip?

(c) What is the average speed for this new trip?

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