Dynamics_Part8

Dynamics_Part8 - Problem Set 1: Problem 7. Problem: Car A...

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Unformatted text preview: Problem Set 1: Problem 7. Problem: Car A begins from rest at time t = 0, with constant acceleration a. Car B is moving in the opposite direction at a constant speed v = -V and, at t = 0, begins constant acceleration 1 a, i.e., 5 = - 1 . The distance between the cars at t = 0 is d. When the cars pass each other, their speeds are 5 equal. (a) Determine the time at which the cars pass each other, , their speed, |v( )| and the distance d. Express your answers in terms of V , a and d. Also, compute x( )/d, where x( ) is distance from Car A's initial position. (b) If d = 500 ft and V = 55 mph, compute a, , |v( )| and x( )/d. . . . ...................... .. ............ . ........................... ................ . ................... .................... ... . . . ... . . .... . .................. . . ........ ... . ... . .. .................... . . . . . . ........ . . . . .. . .... .. . . . ................................................... ............................................... .... .......... .. . . .. ................ ................ ................ .. . . .. .................................... ...... ........ ......... .. ........ .. ... .............................. .... . .. . . .. . ... . ... .. ............................................................... ................ ... .. ....... .... .... ................................ .............................................................................. ............................................................................ .................................................................. . . . .............. .. .. .......................................................................... ..... ................... .......... .. .. . .. .... . .................................................. .................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... .. .............................................................................. .......... . ............. .................................................................................................................................................................................................................................................................................................................................................................................................................................................... ... . ................................. .................................................................................................................................................................................................................................................................................................................................................................................................................................................... . .................. ................. ................. ................. . . v = vi v =v i A B a = ai a = a i x Solution: The equation describing Car A's velocity is dv =a dt dx = v = at dt = v(t) = at where we make use of the fact that the car starts from rest so that v(0) = 0. The car's position is given by = x(t) = 1 2 at 2 where we have chosen the origin to be the point from which Car A begins its motion so that x(0) = 0. Similarly, for Car B, we have dv = a dt = 1 v (t) = -V - at 5 where we use the given facts that v (0) = -V and = - 1 . Car B's position is given by 5 1 dx = v = -V - at dt 5 = x (t) = d - V t - 1 2 at 10 where we use the fact that the distance between the cars at t = 0 is d. (a) Since the cars' speeds are equal when they pass, i.e., v( ) = |v ( )|, we have 1 a = V + a 5 Their positions are also equal, of course, so that = = 5V 4a 1 1 2 a = d - V - a 2 = 2 10 So, substituting for from above, this equation yields d=V 5V 4a 3 + a 5 5V 4a 2 3 d = V + a 2 5 = 35 V 2 5 V 2 15 V 2 + = 4 a 16 a 16 a ...
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