Ratio, rate, and speed (motion graphs)

Advanced Math 7 · Unit 6 · speed-motion-graphs · Teacher edition

speed-motion-graphs

Name ______________________________Date ______________

1.
A train travels at a steady $80$ km/h for $2$ hours $15$ minutes. How far does it travel, in km?
Answer: ______________
2.
Sara drives $60$ km from home to a town at $30$ km/h and returns along the same road at $60$ km/h. What is her average speed for the whole round trip, in km/h?
Answer: ______________
3.
A speed–time graph for a tram: the speed rises in a straight line from $0$ to $20$ m/s during the first $10$ seconds, then stays at $20$ m/s for the next $30$ seconds. How far does the tram travel in these $40$ seconds, in metres?
Answer: ______________
4.
A bus goes from town A to town B at $40$ km/h and returns along the same road at $60$ km/h. What is its average speed for the whole journey?
1. (A)
  $48$ km/h
2. (B)
  $50$ km/h
3. (C)
  $52$ km/h
4. (D)
  $100$ km/h
5.
A walker leaves a village at 8:00 a.m. at a steady $4$ km/h. At 9:30 a.m. a cyclist leaves the same village along the same road at a steady $12$ km/h. At what time does the cyclist catch the walker?
1. (A)
  10:00 a.m.
2. (B)
  10:15 a.m.
3. (C)
  10:30 a.m.
4. (D)
  10:45 a.m.
5. (E)
  11:00 a.m.

Answer key — Ratio, rate, and speed (motion graphs)

1.
180speed-motion-graphs-01
$2$ h $15$ min $= 2.25$ h. Distance $= 80 \times 2.25 = \mathbf{180}$ km. (Using $2.15$ hours is the common slip: minutes are not decimals.)
2.
40speed-motion-graphs-02
Total distance $= 120$ km. Total time $= 2 + 1 = 3$ h. Average speed $= \frac{120}{3} = \mathbf{40}$ km/h — not $45$ , the mean of the two speeds, because twice as long was spent at the slow speed.
3.
700speed-motion-graphs-03
Area $= \frac{1}{2}(10)(20) + 30 \times 20 = 100 + 600 = \mathbf{700}$ m. (Using $20 \times 40 = 800$ ignores that the tram was still speeding up for the first $10$ s.)
4.
(A)
$48$ km/h
speed-motion-graphs-04
With $120$ km each way: $240$ km in $3 + 2 = 5$ h gives $\mathbf{48}$ km/h. ( $50$ is the trap: averaging the speeds is only right when the times are equal. More time is spent at $40$ , so the answer must be below $50$ .)
5.
(B)
10:15 a.m.
speed-motion-graphs-05
Gap at 9:30: $6$ km. Closing speed: $8$ km/h. Time: $\frac{6}{8}$ h $= 45$ min, so the cyclist catches up at 10:15 a.m. Check: walker $2.25 \times 4 = 9$ km; cyclist $0.75 \times 12 = 9$ km. ✓ (10:00 divides the $6$ km gap by $12$ instead of by the closing speed $8$ .)