Kathy tests her new car by racing with Stan. Both start from rest, but Kathy lea
ID: 2077189 • Letter: K
Question
Kathy tests her new car by racing with Stan. Both start from rest, but Kathy leaves the starting line 1.00s after Stan does. Stan moves with constant acceleration of 3.3m/s^2 while Kathy maintains an acceleration of 5.33 m/s^2. Find the time at which Kathy over takes Stan from the time Kathy starts driving. Find the distance she travels before she catches him. Find the speeds of both drivers at the moment she over takes him. Kathy tests her new car by racing with Stan. Both start from rest, but Kathy leaves the starting line 1.00s after Stan does. Stan moves with constant acceleration of 3.3m/s^2 while Kathy maintains an acceleration of 5.33 m/s^2. Find the time at which Kathy over takes Stan from the time Kathy starts driving. Find the distance she travels before she catches him. Find the speeds of both drivers at the moment she over takes him.Explanation / Answer
In 1s, the distance covered by Stan is:
S = ut + (1/2)at2 = 0 + (1/2)3.3(1)2 = 1.65m
and his velocity at this point will be: v = 0 + 3.3(1) = 3.3 m/s
So, in time t, Kathy must cover a distance of, S' = 1.65 + (3.3t + (1/2)(3.3)t2)
S' = 0 + (1/2)(5.33)t2
therefore, 2.665t2 = 1.65 + 3.3t + 1.65t2
=> 1.015t2 - 3.3t - 1.65 = 0
solve this quadratic equation to get:
t = 3.6916s and t = - 0.44s of which the negative value of time must be discarded.
therefore, at time T = 3.6916 + 1 = 4.6916s, Kathy takes over Stan.
for this time, the distance covered by Kathy is: S' = 0 + (1/2)(5.33)(4.6916)2 = 58.660m
At this time, Kathy's speed will be: v = 0 + 5.33(4.6916) = 25 m/s
and Stan's speed will be: v = 0 + 3.3(4.6916) = 15.48 m/s.
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