Two CarsDate: 08/11/99 at 07:42:33 From: Sandi Subject: Kinematics Hi Doctors, I am hoping that you can help me with this problem. Cars A and B are stationary on a straight road, side by side. Car A moves off with acceleration of 1.20 m/(s^2), maintaining this rate until it reaches a speed of 108 km/h (30 m/s). It then continues to cruise at this rate. 15 seconds after A starts, B sets off in pursuit of A with constant acceleration of 2 m/(s^2) until it draws even with A. Find: a) how long car B takes to draw even with car A (to nearest 10th of a second), and b) the distance car B travels, to the nearest meter, before drawing even with car A. So far I have worked out that the displacement of the two cars needs to be the same, so I need two equations that I can say equal each other to solve for t. Car A displacement = [(.5)(25)(30)] + 30(t-25) Car B displacement = [(.5)(15)(30)] + 30(t-45) ... or is it (t-30)? The part that confuses me is that car B starts 15 seconds later. I've tried saying that for car B, V = 2t + 15 then does s = t^2 + 15t + 450 because when v = 0, t = 15. Anyway, I know that the answer is 32.3 seconds and I have tried many different approaches. I tried to make t = 0 when car A reaches 30 m/s but got even more confused. Could you please show me how to do this so that any other questions of this type won't be a problem in the future? Thank you very much, Sandi Date: 08/11/99 at 15:39:34 From: Doctor Anthony Subject: Re: Kinematics >a) how long car B takes to draw even with car A (to nearest 10th of a > second) The time for A to reach 30 m/s is given by v = u + at, so t = (v-u)/a = (30-0)/1.2 = 25 secs and distance is given by s = (1/2)(u+v)t = (1/2)(30)(25) = 375 m Thereafter A will travel 30 m/sec, so if t1 = total time that A has been moving, we know that the distance is s_A = 375 + 30(t1 - 25) ...........................(1) The total time that B moves until it reaches A is t1-15. The distance traveled by B is s_B = 0 + (1/2)(2)(t1-15)^2 .......................(2) Equating the distances in (1) and (2) (t1-15)^2 = 375 + 30(t1-25) (t1-15)^2 - 30(t1-25) - 375 = 0 t1^2 - 30.t1 + 225 - 30.t1 +750 -375 = 0 t1^2 - 60.t1 + 600 = 0 This is a quadratic in t1, which has a positive root 30+10.sqrt(3), and so t1 = 30 + 10.sqrt(3) = 47.32 sec. or 12.68 sec. and this is the time A has been moving, so the time for B to overtake A (from when B starts) is t1 - 15 = 47.32 - 15 sec. or 12.68 - 15 sec. = 32.32 sec. or -2.32 sec. We can disregard the latter, and our answer is 32.32 sec. >b) the distance car B travels, to the nearest meter, before drawing > even with car A. This is the same distance as car A has traveled s = 375 + 30(t1 - 25) = 375 + 30[30+10.sqrt(3)-25] = 375 + 884.847 = 1044.615 meters - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/