p0224 - 2.24. CHAPTER 2, PROBLEM 24 119 2.24 Chapter 2,...

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Unformatted text preview: 2.24. CHAPTER 2, PROBLEM 24 119 2.24 Chapter 2, Problem 24 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 λ = 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, τ , their speed, |v(τ )|, and the distance d. Express your answers in terms of V and a. Also, compute x(τ )/d, where x(τ ) is distance from Car A’s initial position. (b) If V = 100 km/hr and the cars pass each other when τ = 6 sec, compute a, d, |v(τ )| and x(τ ). v = vi A v =v i B x . . . ................ ............... ................ ............... . . ................ . ............................................................................................................................................................................................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................................................................................................................................................................................... . . . . ................. ................. ................. ................. a = ai a = λa i 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) = 12 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 = λa dt =⇒ 1 v (t) = −V + at 3 where we use the given facts that v (0) = −V and λ = 1 . Car B’s position is given by 3 1 dx = 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 (τ ) = |v (τ )|, we have 1 aτ = V − aτ 3 Their positions are also equal, of course, so that =⇒ τ= 3V 4a 12 1 =⇒ aτ = d − V τ + aτ 2 2 6 So, substituting for τ from above, this equation yields d=V 3V 4a 1 +a 3 3V 4a 2 1 d = V τ + aτ 2 3 = 3 V2 15 V 2 3V2 + = 4a 16 a 16 a 120 The speed of the cars, |v(τ )| = v (τ ), is thus |v(τ )| = a 3V 4a = CHAPTER 2. PARTICLE KINEMATICS 3 V 4 Finally, the point at which the cars pass each other is x(τ ) = Therefore, we find x(τ ) = d 9 2 32 V /a 15 2 16 V /a 121 aτ = a 2 2 3V 4a 2 = 9 V2 32 a = 3 10 Summarizing, details of the results obtained are as follows. τ= 3V , 4a d= 15 V 2 , 16 a |v(τ )| = 3 V, 4 x(τ ) 3 = d 10 (b) For the given data, the numerical values of a, |v(τ )| and x(τ ) are 3V 3 a= = 4τ 4 15 V 2 15 = 16 a 16 100 km hr km hr 1 hr 3600 sec 6 sec 1000 m km = 3.47 2 m sec2 100 d= 1 hr m 1000 3600 sec km 3.47 m/sec2 100 km hr = 75 km hr = 208.5 m |v(τ )| = x(τ ) = 3 3 V= 4 4 3 3 d= (208.5 m) = 62.6 m 10 10 ...
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