Mary and Sally are in a foot race (see figure (Figure 1) ). When Mary is 22 m fr
ID: 1606140 • 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 0.48 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?
Mary Sally 4.0 m/s 5.0 m/s 22 m FinishExplanation / Answer
Given:
Mary Sally
Displacement 22m 17m
Initial Velocity 4m/s 5m/s
Final Velocity unknown
Acceleration -0.5m/s^2
Time same
Time for Sally:
D= v1(t)-1/2a(t)^2
17m=5m/s(t)+0.125(t)^2
Use the quadratic formula to find both values for t which is
3.16s or -43.16s
use 3.16 cause time can't be negative
since time for sally and mary are the same,
we find acceleration for mary:
D= v1(t)+1/2a(t)^2
22m=4m/s(3.16s)+1/2(a)(3.16)^2
1/2(a)(3.16)^2=9.36
a=1.872m/s^2
Mary must accelerate at 1.872m/s^2 to cross the finish line side by side with Sally.