Question

. Mary and Sally are in a foot race (Fig. 2). When Mary is 22 m from the finish line, she has a speed of 4.0 m/s and is 5.0 m behind Sally, who has a speed of 5.0 m/s. Sally thinks she has an easy win and so, during the remaining portion of the race, decelerates at a constant rate of 0.4 m/s2 to the finish line. What constant acceleration does Mary now need during the remaining portion of the race, if she wishes to cross the finish line sideby-side with Sally?

Answer 1

Answer on Question #65777, Physics / Mechanics | Relativity

Mary and Sally are in a foot race (Fig. 2). When Mary is 22 m from the finish line, she has a speed of 4.0 m/s and is 5.0 m behind Sally, who has a speed of 5.0 m/s. Sally thinks she has an easy win and so, during the remaining portion of the race, decelerates at a constant rate of 0.4 m/s² to the finish line. What constant acceleration does Mary now need during the remaining portion of the race, if she wishes to cross the finish line side by-side with Sally?

Solution:

Find how long it takes Sally to reach the finish line:

$$x = v _ {0} t + \frac {1}{2} a t ^ {2}$$ $$(22 - 5) m = (5.0 \, m/s)(t) + (1/2)(-0.4 \, m/s^2)t^2$$ $$0.2t^2 - 5t + 17 = 0$$ $$t = \frac {5 \pm \sqrt {(-5)^2 - 4(0.2)(17)}}{2 \cdot (0.2)} = 12.5 \pm 8.44 =$$

Solve for t.

t = 20.94 or 4.06 seconds

t = 4.06 seconds.

We want the smaller answer – the other answer is the time it would take to pass the finish line, change direction, and cross the finish line again.

$$t = 4.06 \, s$$

What constant acceleration does Mary now need during the remaining portion of the race, if she wishes to cross the finish line side-by-side with Sally?

Since time for Sally equals times for Mary. Reuse t = 4.06 s.

Find the acceleration Mary would need.

$$x = v _ {0} t + \frac {1}{2} a t ^ {2}$$ $$22 \, m = (4.0 \, m/s)(4.06 \, s) + (1/2)(a)(4.06 \, s)^2$$ $$a = \frac {22 - 16.24}{0.5 \cdot 4.06^2} = 0.7 \, m/s^2$$

Answer: $0.7 \, m/s^2$

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