Question

Finding Initial Value and Rate of Change
Bus Stop: A graph has time (minutes) on the x-axis, and distance (miles) on the y-axis. A line goes through points (0, 5), (6, 8), and (12, 11). Friend's house: A 2-column table with 3 rows. Column 1 is labeled Time (minutes) with entries 3, 6, 9. Column 2 is labeled Distance (miles) with entries 4, 6, 8.

Seth is comparing scenarios to determine which starting point he should get dropped off at to get him to school faster. Seth's house is 17 miles from school.



In which scenario does Seth start closer to school?


In which scenario does Seth travel at a greater speed, and what was the rate?

Answer 1

Find the initial distances (time 0) and the rates (slopes).

Bus stop:
- Points (0,5), (6,8) ⇒ slope = (8−5)/(6−0) = 3/6 = 0.5 miles/min.
- Initial distance at t=0 = 5 miles.
(So equation: distance = 5 + 0.5 t.)

Friend's house:
- Points (3,4), (6,6) ⇒ slope = (6−4)/(6−3) = 2/3 miles/min.
- Extrapolate to t=0: distance = 4 − (2/3)·3 = 2 miles.
(So equation: distance = 2 + (2/3) t.)

Answers:
- Seth starts closer to school at the friend’s house (2 miles) rather than the bus stop (5 miles).
- The friend’s-house scenario has the greater speed: 2/3 mile per minute (≈0.6667 mi/min), which is 40 mi/h. The bus speed is 0.5 mi/min (30 mi/h).