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 0.49 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.

Answer 1

The expression of the kinematic equation for Sally is, v2 – vě = 2a,s Here, v is the final speed of Sally, V, is the initial speed of Sally, ag is the deceleration of Sally, and s is the distance travelled by Sally. The range of the foot race is 22m and the distance between Sally and Mary is 5 m. Then the distance travelled by Sally is, S = 22 m - 5m = 17 m Sally decelerates from her initial speed to rest, then the acceleration of Sally should have a negative sign to represent that she is decelerating. Ag =-a Calculate the final speed of Sally as follows: 12 – Vă = 2ags v= 12(-a) s + v = V-(2)(0.49 m/s?)(17m)+(5.0m/s)? = 2.888m/s The expression of the kinematic equation for Sally is, v=v+agt

Calculate the time taken for Sally to complete the foot race as follows: v=Vo +act V-V t= ad v-V -a 2.888 m/s -5.0 m/s -0.49 m/s2 = 4.33 The time taken for Sally is must equal to time taken for Mary, because Mary wants to finish the race side by side with Sally. The expression of the kinematic equation for Mary is, s'= våt + alt? Here, v is the initial speed of Sally, d' is the acceleration of Mary, and s' is the distance travelled by Mary.

Calculate the acceleration of Mary as follows: s'= vot + at d = 2(s'– výt) _(2)(22m-(4.0 m/s)(4.3s)) (4.3s) = 0.52 m/s2 Therefore, the acceleration of Mary is 0.52 m/s?.