200 m In a 400-m race, runer A reaches her maxinmum velocity v in 4 s with const
ID: 2033909 • Letter: 2
Question
200 m In a 400-m race, runer A reaches her maxinmum velocity v in 4 s with constant acceleration and maintains that velocity until she reaches the half-way point with a split time of 25 s. Runner B reaches her maxinmum velocity v in 5 s with constant acceleration and maintains that velocity until she reaches the half-way point with a split time of 25.2 s. Both runners then run the second half of the race with the same constant deceleration of 0.1 m/s2. Determine (a) the race times for both rumners, (b) the position of the winner relative to the loser when the winner reaches the finish line.Explanation / Answer
for A:
vf = vi + a t
vA = = 0 + 4 a1
vA = 4 a1
distane travelled during this time,
d= v0 t + a t^2 / 2
d1 = 0 + (a1)(4^2)/2 = 8 a1
after that A run with constant velocity 4a1 for 21 s.
the d2 = (4a1)(21) = 84 a1
d1 +d2 = 200
8a1 + 84 a1 = 200
a1 = 2.174 m/s^2
for B:
vf = vi + a t
vB = = 0 + 5 a2
vB = 5 a2
distane travelled during this time,
d= v0 t + a t^2 / 2
d1 = 0 + (a2)(5^2)/2 = 12.5 a1
after that A run with constant velocity 4a1 for 21 s.
the d2 = (5a2)(25.2 - 5) = 101a2
d1 +d2 = 200
12.5a2 + 101 a2 = 200
a2 = 1.762 m/s^2
(A) for A:
vA = 4 a1 = 8.696 m/s
200 = (8.696 t1)+ ( - 0.1 t1^2 /2)
0.05 t1^2 - 8.696 t + 200 = 0
t1 = 27.3 sec
race time = 25 + 27.3 = 52.3 sec
for B:
vB = 5 a2 = 8.81 m/s
then 200 = 8.81 t - 0.1 t^2 /2
t = 26.8 sec
race time = 26.8 + 25.2 = 52 s
(B) Winner is B.
so position of A at 52 s.
t1 = 52 - 25 = 27 sec
d3 = (8.696)(27)+(- 0.1 x 27^2 / 2) = 198.3 m
position = 200 + 198.3 = 398.3 m
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