Courier Distance Function: Average Rate, Two Solutions, Invertibility
A courier van's distance from its depot is logged continuously from 8:00 to 9:30. From 8:00 to 8:30 the van drives outbound at a steady 0.6 miles per minute. From 8:30 to 8:45 it is parked at a delivery. From 8:45 to 9:15 it runs a second outbound leg at a steady 0.8 miles per minute. From 9:15 to 9:30 it heads back toward the depot at a steady 2 miles per minute. Let D(t) be the distance in miles and t the number of minutes after 8:00; the logged checkpoints are D(0) = 0, D(30) = 18, D(45) = 18, D(75) = 42, and D(90) = 12. The dispatcher has three questions: how fast the van is progressing on average across the whole window, at what clock times the van is exactly 30 miles from the depot, and whether D can be inverted to answer 'when was the van m miles out?' for every distance the van actually reached.
1. Compute the average rate of change of D over the full 90 minutes, and compare it with the rate on each segment. Explain the discrepancy.
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Average rate over [0, 90]: (D(90) − D(0)) / (90 − 0) = (12 − 0) / 90 = 0.1333 miles per minute, i.e. about 8.0 mph (0.1333 × 60).
Segment rates:
- [0, 30]: (18 − 0)/30 = 0.6 mi/min (36 mph)
- [30, 45]: (18 − 18)/15 = 0 mi/min (parked)
- [45, 75]: (42 − 18)/30 = 0.8 mi/min (48 mph)
- [75, 90]: (12 − 42)/15 = −2 mi/min (returning at 120 mph)
Why they differ: the average rate is a net displacement over total time. The van ends only 12 miles from the depot, so the 30 miles of outward travel in the last leg are cancelled by the return. The average is far slower than any driving speed because it absorbs a 15-minute stop and 15 minutes of travel back toward the origin. Reporting 8 mph as 'the van's speed' would be wrong; it is a net progress figure, useful for ETA-to-endpoint questions only.
2. Solve D(t) = 30. Give every solution in minutes after 8:00 and in clock time, and justify why no other solutions exist.
Read the worked answer
Check each segment's range:
- [0, 30]: D rises 0 → 18. Never reaches 30.
- [30, 45]: D is constant at 18. Not 30.
- [45, 75]: D rises 18 → 42, so 30 is attained once.
- [75, 90]: D falls 42 → 12, so 30 is attained once.
Segment 3: D = 18 + 0.8(t − 45). Set 18 + 0.8(t − 45) = 30 → 0.8(t − 45) = 12 → t − 45 = 15 → t = 60, i.e. 9:00.
Segment 4: D = 42 − 2(t − 75). Set 42 − 2(t − 75) = 30 → 2(t − 75) = 12 → t − 75 = 6 → t = 81, i.e. 9:21.
Verification: D(60) = 18 + 0.8(15) = 30 ✓ and D(81) = 42 − 2(6) = 30 ✓.
Why only two: on each linear segment D is monotonic, so it can cross the value 30 at most once, and segments 1 and 2 never reach 30. Two crossings total. The dispatcher must not report a single time - the van was 30 miles out twice, outbound and again on the return.
3. Determine whether D has an inverse on [0, 90]. If not, restrict the domain to give a valid inverse for the outbound leg and state its domain and range.
Read the worked answer
No inverse on the full domain. Invertibility requires one-to-one behaviour, and the previous task shows D(60) = D(81) = 30: the same output comes from two inputs. The function is not injective, so 't = D⁻¹(m)' would be ambiguous.
Restrict to [45, 75], where D is strictly increasing from 18 to 42: D = 18 + 0.8(t − 45) D − 18 = 0.8(t − 45) t − 45 = 1.25(D − 18) D⁻¹(m) = 45 + 1.25(m − 18), with domain 18 ≤ m ≤ 42 and range 45 ≤ D⁻¹ ≤ 75.
Check: m = 30 → D⁻¹ = 45 + 1.25(12) = 60 ✓, matching the outbound solution.
Alternative valid restriction: [75, 90] gives t = 75 + (42 − m)/2 for 12 ≤ m ≤ 42, which returns the return-leg times (m = 30 → 81). Because several restrictions are valid, the dispatcher's question must always specify which leg is meant. The parked interval [30, 45] cannot be used at all: constant D = 18 is not one-to-one.