Question

A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 15.0 min at 90.0 km/h, 7.0 min at 65.0 km/h, and 40.0 min at 40.0 km/h and spends 40.0 min eating lunch and buying gas.(a) Determine the average speed for the trip. km/h(b) Determine the distance between the initial and final cities along the route. km

Answer 1

a)

To determine the average speed we need to determine the time the person travels and the distance. Since we need the average speed in km/h we need to convert all the times given into hours, let's do this:

`\begin{gathered} 15\text{ min}\cdot\frac{1\text{ h}}{60\text{ min}}=0.25\text{ h} \\ 7\text{ min}\cdot\frac{1\text{ h}}{60\text{ min}}=(7)/(60)\text{ h} \\ 40\text{ min}\cdot\frac{1\text{ h}}{60\text{ min}}=(2)/(3)\text{ h} \end{gathered}`

Now, that we have the time in each interval of the motion we can determine the distance is travel in each of them:

`\begin{gathered} d_1=90(0.25)=22.5 \\ d_2=65((7)/(60))=7.583 \\ d_3=40((2)/(3))=26.667 \end{gathered}`

This means that the total distance the person traveled is 56.8 km (we use three significant figures as shown in the problem).

Once we know this we need to find the time it takes to make this, from our previos discussion and adding the 40 minutes the person took to eat we have that the total time was 1.7 h, then the average speed is:

`v=(56.8)/(1.7)=33.4`

Therefore, the average speed on the trip was 33.4 km/h

b)

From point a we know that the distance was 56.8 km