3. Jane is walking her dog, Spot. She sees her friend, Dick, walking toward her

Question

3. Jane is walking her dog, Spot. She sees her friend, Dick, walking toward her along the same long, straight road. Both Dick and Jane are walking at 3 mph. When Dick and Jane are 600 feet apart, Spot runs from Dick to Jane, turns and runs back to Dick, and then back and forth between them at a constant speed of 8 mph. Dick and Jane both continue walking toward each other at a constant 3 mph. Neglecting the time lost each time Spot reverses direction, how far has Spot run in the time it takes Dick and Jane to meet?

Explanation / Answer

Dick and Jane are initially 600 feet apart. Each walks half that distance, or 300 feet, to meet. This takes t = d/v = 300/3 = 100 time units. We are using "mixed" units here, distance in feet, speed in miles/hour, and time in (foot-hours)/mile. But who cares about the time units so long as we are consistent? In that time the dog travels d = vt = 8×100 = 800 feet.

Using mixed units avoids a unit conversion from miles/hour to ft/second, making the computation easier to do in one's head. We weren't asked to find the time, so it drops out of the calculation. In fact, a better approach is to do the algebra first, then insert numbers, which is nearly always a good strategy. We also weren't asked to calculate the ever-decreasing distances for each of Spot's runs between Dick and Jane. t = d/v where d is the distance Dick and Jane walk, and v is their speed.

D = Vt where D is the distance Spot runs and V is Spot's speed.

Then D = Vd/v = 8×300/3 = 800 ft.

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