Calculating Average Velocity: The Subway Train

What is the average velocity of the train in part b of Example 2.2, and shown again below, if it takes 5.00 min to make its trip?

Figure 2.22

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
Average velocity is displacement divided by time. It will be negative here, since the train moves to the left and has a negative displacement.
Solution
1. Identify the knowns. ${x'}_{f} = 3 .75 km$ , ${x'}_{0} = 5.25 km$ , $Δ t = 5.00 min$ .
2. Determine displacement, $Δ x'$ . We found $Δ x'$ to be $- 1.5 km$ in Example 2.2.
3. Solve for average velocity.

$\bar{v} = \frac{Δ x'}{Δ t} = \frac{- 1.50 km}{5.00 min}$
2.19

4. Convert units.

$\bar{v} = \frac{Δ x'}{Δ t} = (\frac{- 1 . 50 km}{5 . 00 min}) (\frac{60 min}{1 h}) = - 18 .0 km/h$
2.20
Discussion
The negative velocity indicates motion to the left.

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