Question

Calculate the average speed and average velocity of a complete round trip in which the outgoing 250 km is covered at 95 km/h, followed by a 1.0 h launch break, and the return of 250 km is covered at 55 km/h.

Short answer

The average speed for this round trip is $61.12\frac{\mathrm{km}}{\mathrm{h}}$, and the average velocity for this trip is 0.

Step-by-step solution

Step 1. The relation between the average speed, the average velocity with distance, displacement, and time

The average speed of an object is defined as the average distance covered by an object in unit time.

Given data.

For the first part of the trip, 250 km is covered at $95\frac{\mathrm{km}}{\mathrm{h}}$.

For the second part of the trip, there is a break for 1.0 h.

For the third part of the trip, 250 km is covered at $55\frac{\mathrm{km}}{\mathrm{h}}$.

Now, you know that

$\begin{array}{c}\mathrm{Average}\mathrm{speed}=\frac{\mathrm{Total}\mathrm{distance}}{\mathrm{Total}\mathrm{time}}\\ \mathrm{Average}\mathrm{velocity}=\frac{\mathrm{Total}\mathrm{displacement}}{\mathrm{Total}\mathrm{time}}\end{array}$

Assumptions.

Let the time taken for the first part of the trip be $t_1\mathrm{h}$, and that for the third part of the trip be $t_3\mathrm{h}$.

Step 2. Calculation of the total distance, total displacement, and total time

Here, the total distance is the sum of the distances in three parts.

$\begin{array}{c}\mathrm{d}=\mathrm{distance}\mathrm{for}\mathrm{first}\mathrm{part}+\mathrm{distance}\mathrm{for}\mathrm{the}\mathrm{second}\mathrm{part}+\mathrm{distance}\mathrm{for}\mathrm{second}\mathrm{part}\\ =250\mathrm{km}+0+250\mathrm{km}\\ =500\mathrm{km}\end{array}$

For the complete round trip, the displacement is zero because the initial and final points are the same.

Now, the total time is

$\begin{array}{c}\Delta t=t_1+t_2+t_3\\ =\frac{250\mathrm{km}}{95\frac{\mathrm{km}}{\mathrm{h}}}+1.0\mathrm{h}+\frac{250\mathrm{km}}{55\frac{\mathrm{km}}{\mathrm{h}}}\\ =8.18\mathrm{h}\end{array}$

Step 3. Calculation of average speed and average velocity

Therefore,

$\begin{array}{c}\mathrm{Average}\mathrm{speed}=\frac{500\mathrm{km}}{8.18\mathrm{h}}\\ =61.12\frac{\mathrm{km}}{\mathrm{h}}\\ \mathrm{Average}\mathrm{velocity}=\frac{0\mathrm{km}}{8.18\mathrm{h}}\\ =0\end{array}$

Hence, the average speed for this round trip is $61.12\frac{\mathrm{km}}{\mathrm{h}},$and the average velocity for this trip is 0.