PS1-8

# PS1-8 - Problem Set 1: Problem 8. Problem: Car A begins...

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Unformatted text preview: Problem Set 1: Problem 8. 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 deceleration 1 a, i.e., 3 = 1 . The distance between the cars at t = 0 is d. When the cars pass each other, their speeds are 3 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 = 200 m and V = 100 km/hr and the cars pass each other when = 6 sec, 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 = 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 dx = v = at dt = 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 3 where we use the given facts that v (0) = -V and = 1 . Car B's position is given by 3 dx 1 = v = -V + at dt 3 = 1 x (t) = d - V t + at2 6 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 3 Their positions are also equal, of course, so that 1 1 2 a = d - V + a 2 2 6 = 1 d = V + a 2 3 = = 3V 4a So, substituting for from above, this equation yields d=V 3V 4a 1 + a 3 3V 4a 2 = 3V2 3 V2 15 V 2 + = 4 a 16 a 16 a The speed of the cars, |v( )| = v( ), is thus |v( )| = a 3V 4a = 3 V 4 Finally, the point at which the cars pass each other is x( ) = Therefore, we find x( ) = d 9 2 32 V /a 15 2 16 V /a 1 2 1 a = a 2 2 3V 4a 2 = 9 V2 32 a = 3 10 Summarizing, details of the results obtained are as follows. = 3V , 4a d= 15 V 2 , 16 a |v( )| = 3 V, 4 x( ) 3 = d 10 (b) For the given data, the numerical values of a, |v( )| and x( ) are 3V 3 a= = 4 4 100 km hr 1 hr 3600 sec 6 sec 100 km hr 1000 m km km hr = 3.47 m sec2 |v( )| = 3 3 V = 4 4 = 75 x( ) = 3 3 d= (200 m) = 60 m 10 10 ...
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## This note was uploaded on 10/12/2009 for the course AME 301 at USC.

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