Average Velocity and Average Speed — Question 6

Question 6

Can a later leg recover a target average?

A traveler follows a 90 km straight eastward route without reversing. The first 30 km take 1.0 h, followed by a 0.25 h stop. The remaining 60 km are to be traveled at a steady positive speed $u$. The trip average includes the stop. Treat the route and times as exact model values; ignore time needed to change speed.

1. Find the value of $u$ needed for a full-trip average speed of $45\ \mathrm{km/h}$. What is the corresponding average velocity on an east-positive axis?

2. Derive $u$ as a function of a proposed full-trip average speed $S$. State precisely which positive $S$ values are possible with finite positive $u$ when no maximum speed is imposed.

3. Explain why $S=72\ \mathrm{km/h}$ cannot be reached with any finite $u$, and why larger targets are impossible even in the idealized model.

4. Now impose the additional constraint $u\leq80\ \mathrm{km/h}$. Find the greatest attainable full-trip average speed and decide whether the 45 km/h target remains feasible.

Question 6 - Solution

Can a later leg recover a target average?

(a) Convert the desired average into a total time allowance.

The target permits $90/45=2.00\ \mathrm{h}$ overall. The first leg and stop already use $1.25\ \mathrm{h}$, leaving $0.75\ \mathrm{h}$ for 60 km. Thus $$\boxed{u=60/0.75=80\ \mathrm{km/h}}.$$ The motion is eastward without reversal, so $\boxed{\bar v=+45\ \mathrm{km/h}}$.

(b) Remaining time must be strictly positive.

Writing $S$ and $u$ numerically in km/h, the required duration in hours is $90/S$. Hence $$\boxed{u=\frac{60}{90/S-1.25}\ \mathrm{km/h}}.$$ Finite positive $u$ requires $90/S>1.25$ with $S>0$, giving $$\boxed{0<S<72\ \mathrm{km/h}}.$$ Conversely, each target in this open interval gives a finite positive denominator and a valid speed, so the condition is sufficient as well as necessary.

(c) A spent time budget cannot be recovered.

$$S(u)=\frac{90}{1.25+60/u}\ \mathrm{km/h}<72\ \mathrm{km/h},\qquad\lim_{u\to\infty}S(u)=72\ \mathrm{km/h}.$$

At 72 km/h the allowed total time is exactly 1.25 h, all of which has already elapsed. The remaining 60 km would have to take zero time, requiring an unbounded speed. A larger target permits even less than the time already spent. The value 72 km/h is a limiting upper value as $u$ grows, not an attainable maximum at finite speed.

(d) A finite speed limit gives an attainable maximum.

The shortest remaining time with $u\leq80\ \mathrm{km/h}$ is $60/80=0.75\ \mathrm{h}$. Thus total time is at least 2 h and $$\boxed{\bar s_{\max}=90/2=45\ \mathrm{km/h}}.$$ The target remains feasible, exactly at the limit $u=80\ \mathrm{km/h}$. Averaging 30 and 80 km/h arithmetically would ignore both the unequal stage times and the stop.