Question

Julia can run a mile race in 4.25 min. Jasmine requires 4.55 min to run this distance. If they start out together and maintain their normal speeds, how far apart will they be at the finish of the race?

Answer 1

Answer on Question #53505, Physics / Mechanics | Kinematics | Dynamics

Julia can run a mile race in 4.25 min. Jasmine requires 4.55 min to run this distance. If they start out together and maintain their normal speeds, how far apart will they be at the finish of the race?

Solution:

The speed of Julia is

$$v_1 = \frac{s}{t_1} = \frac{1 \text{ mile}}{4.25 \text{ min}} = 0.23529 \text{ mi/min}$$

The speed of Jasmine is

$$v_2 = \frac{s}{t_2} = \frac{1 \text{ mile}}{4.55 \text{ min}} = 0.21978 \text{ mi/min}$$

If the two runners will start together in a mile race, then Julia will finish the race. Therefore, their distance between the two runners at the end of the race is

$$d = s_1 - s_2 = 1 \text{ mile} - v_2 t_1$$ $$d = 1 - 0.21978 * 4.25 = 0.065935 \text{ miles}$$

or

$$d = 0.065935 \text{ miles} * \frac{5280 \text{ ft}}{1 \text{ mile}} = 348.14 \text{ ft}$$

or

$$d = 348.14 \text{ ft} * \frac{0.3048 \text{ m}}{1 \text{ m}} = 106.11 \text{ m}$$

Answer: $d \approx 348 \text{ ft} \approx 106 \text{ m}$

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