21. o a10 points 1 Previous Answers SerPSE9 2 POT3 MI Netes Ask Your Kathy tests

Question

21. o a10 points 1 Previous Answers SerPSE9 2 POT3 MI Netes Ask Your Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting ine 1.00 s after Stan does. Stan moves with a constant acceleration of 3.6 m/s2 while Kathy maintains an acceleration of 4.49 mýs? e) Find the time at which Kathy overtakes Stan. differs from the correct answer by more than 10%. Double check your calculations. 5 from the time Kathy started driving (b) Find the distance she travels before she catches him. (c) Find the speeds of beth cars at the instant she overtakes him. Kathy Stan my's m/s

Explanation / Answer

For Kathy, the time is t – 1. Use the following equation to determine the distance she moves.

d = vi * t + ½ * a * t^2, vi = 0
d = ½ * 4.49 * (t – 1)^2
(t – 1)^2 = t^2 – 2 * t + 1
d = 2.245 * t^2 – 4.49 * t + 2.245

For Stan, d = ½ * 3.6 * t^2 = 1.8 * t^2
Set these two equations equal to each other and solve for.

2.245 * t^2 – 4.49 * t + 2.245 = 1.8 * t^2
0.445 * t^2 – 4.49 * t + 2.245 = 0
t = [4.49 ± (-4.49^2 – 4 * 0.445 * 2.245)] ÷ 0.89
This is approximately 9.56 seconds.
other value
This is approximately 0.52 seconds. Since this is less than one second, this is not the correct answer.

(b) Find the distance she travels before she catches him.

d = 2.245 * 9.56^2 – 4.49 * 9.56+ 2.245
This is approximately 164.5 meters

(c) Find the speeds of both cars at the instant she overtakes him.

For Kathy, the time is (4.49 + 16.164) ÷ (0.89) – 1
This is approximately 8.56 seconds

vf = vi + a * t, vi = 0
vf = 4.49 * (4.49 + 16.164) ÷ (0.89) – 1
This is approximately 38.44 m/s.

For Stan, vf = 3.6 *(4.49 + 16.164) ÷ (0.89)
This is approximately 34.42 m/s.

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