Mary and Sally are in a foot race (see figure (Figure 1) ). When Mary is 22 m fr
ID: 1605929 • Letter: M
Question
Mary and Sally are in a foot race (see figure (Figure 1) ). 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 0.46 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?
Express your answer using two significant figures.
Figure 1 of 1 Mary Sally 4.0 m/s 5.0 m/s 22 m FinishExplanation / Answer
time= speed/ disatnce
4 t + 1/2 at^2 = 22-----------Mary--eq(1)
5t - 1/2 0.46t^2 = 17--------Sally----eq(2)
solvibg eq(2) further , we get
5t - 0.23t^2 = 17
0.23t^2 -5t + 17=0
which gives, t = 5 +- sqroot ( 25 - 15.64)/ 2 x 0.23 = 4.218 sec apprx
4( 4.218 ) + 0.5( 4.218)^2 a = 22
a= 0.576 m/s ^2 apprx