**Problem1A-5*:** One-Dimensional Rollercoaster

In a simple one-dimensional roller coaster, the car leaves the station, moves along a curved course up to a turn-around point, reverses its direction, and returns to the station by back-tracking along the course. The motion of the car is described by the function

f(t) = Atwhere f(t) is the distance of the car from the station, measured along the curve of the course, and A, B and C are given constants with values^{3}+ Bt^{2}+ Ct ,

A = 1.00×10^{-3}m/sec^{3}, B = -6.00×10^{-1}m/sec^{2}, C = 9.00×10 m/sec.

(a) How far is the car from the station at t = 1 min, and at t = 3 min, resapectively?

(b) What are the speeds of the car at t = 1 min and t = 3 min, respectively?

(c) When and where will the car reverse its direction?

(d) How long does the whole ride last?

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**Solution:**

(a) 1 min=60 sec, and 3 min=180 sec. Thus

f(60 sec) = (1.00×10Note: f(180 sec) ‹ f(60 sec) means that the car has already reversed direction at t=180 sec.^{-3}m/sec^{3})×(60 sec)^{3}-(6.00×10^{-1}m/sec^{2})×(60 sec)^{2}+ (9.00×10 m/sec)×(60 sec) = 3.46×10^{3}m . f(180 sec) = (1.00×10^{-3}m/sec^{3})×(180 sec)^{3}-(6.00×10^{-1}m/sec^{2})×(180 sec)^{2}+ (9.00×10 m/sec)×(180 sec) = 2.59×10^{3}m.

(b) The speed of the car, v(t), at time t is

Thus v(60 sec) = 3×(1.00×10^{-3}m/sec^{3})×(60 sec)^{2}-2×(6.00×10^{-1})×(60 sec) + (9.00×10 m/sec) = 2.88×10 m/sec. v(180 sec) = 3×(1.00×10^{-3}m/sec^{3})×(180 sec)^{2}-2×(6.00×10^{-1})×(180 sec) + (9.00×10 m/sec) = -2.88×10 m/sec. Note: the minus sign in front of the answer of v(180 sec) means that the car has reversed the direction already and the car is now moving toward the station.

(c) The reversal of the direction of the car occurs at v(t) = 0. Thus solving

Since B^{2}-3AC = (-6.00×10^{-1})2-3×(1.00×10^{-3})×(9.00×10) = (0.300)^{2}, the solution is t = 1.00×10^{2}sec or t = 3.00×10^{2}sec. To see which one is the desirable solution we need to consider the acceleration a(t) = dv(t)/dt = 6At + 2B . We will have a(100 sec) = 6×(1.00×10^{-3}m/sec^{3})×(100 sec) + 2×(6.00×10^{-1}m/sec) = -0.6 m/sec^{2}. This means that at t=100 sec the speed becomes zero but the acceleration becomes negative so that the car will be pulled back toward the station. Thus t=1.00×10^{2}sec is the answer. Since f(100 sec) = (1.00×10^{-3}m/sec^{3})×(100 sec)^{3}- (6.00×10^{-1}m/sec^{2})×(100 sec)^{2}+ (9.00×10 m/sec)×100 sec) = 4.00×10^{3}m, The reversal of the direction of the car occurs at 4.00×10^{3}m from the station at time t=1.00×10^{2}sec.

(d) When the car returns to the station, f(t) becomes zero again. Thus we need to solve the equation f (t) = 0. This leads to

f(t) = At^{3}+ Bt^{2}+ Ct = t(At^{2}+ Bt + C) = 0. Since t = 0 is for the car at the moment to leave the station, we have Now B^{2}-4AC = (-6.00×10^{-1})^{2}-4×(1.00×10^{-3})×(9.00×10) = 0.360-0.360 = 0. Thus the ride takes t = -B/2A = 3.00×10^{2}seconds. Note: In (c) another solution for v(t) = 0 is t = 300 seconds. This means that in this case the speed of the car becomes zero when it returns to the station, but in general this condition does not need to be true.

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