Question #263677

Mary and Sally are in a foot race (see figure (Figure 1)). When Mary is 22 m

m from the finish line, she has a speed of 4.0 m/s

m/s and is 5.0 m

m behind Sally, who has a speed of 5.0 m/s

m/s. Sally thinks she has an easy win and so, during the remaining portion of the race, decelerates at a constant rate of 0.43 m/s

2

m/s2 to the finish line.



Part A

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?


1
Expert's answer
2021-11-10T10:09:16-0500

{sM=vMt+aMt22,sS=vStaSt22;\begin{cases} s_M=v_Mt+\frac{a_Mt^2}2, \\ s_S=v_St-\frac{a_St^2}2; \end{cases}


{22=4t+aMt22,17=5t0.43t22;\begin{cases} 22=4t+\frac{a_Mt^2}2, \\ 17=5t-\frac{0.43t^2}2; \end{cases}     \implies

aM=0.64 ms2.a_M=0.64~\frac m{s^2}.


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Comments

Efe
10.11.21, 18:23

Thats nice thanks for your attachment

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