A car in a roller coaster moves along a track that consists of a sequence of ups

Question

A car in a roller coaster moves along a track that consists of a sequence of ups and downs. Let the x axis be parallel to the ground and the positive y axis point upward. In the time interval from t=0 to t=4s, the trajectory of the car along a certain section of the track is given by
r=A(1m/s)ti+A[(1m/s^3)t^(3)-6(1m/s^2)t^2]j
where A is a positive dimensionless constant.

-derive a general expression for the speed v of the car?
-the roller coaster is designed according to safety regulations that prohibit the speed of the carfrom exceeding 20m/s. Find the maximum value of A allowed by these regulations?

Explanation / Answer

r=A(1m/s)ti+A[(1m/s^3)t^(3)-6(1m/s^2)t^2]j =>v = dr/dt = A(1m/s)i+A[(1m/s^3)3t^(2)-12(1m/s^2)t]j =>Speed = sqrt(A^2 + A^2(3t^2-12t)^2) 20 = Asqrt(1 + (3t^2-12t)^2) A = 20/sqrt(1 + (3t^2-12t)^2) A to be maximum f = (3t^2-12t)^2 has to be minimum. =>f' = 2(3t^2-12t)(6t-12) = 0 =>2t(3t-12)(6t-12) = 0 =>t = 0, 4, 2 f'' = 2(3t-12)(6t-12)+6t(6t-12)+12t(3t-12) f''(0) = 288>0.....Minimum f''(2) = -144Amax = 20/1 = 20

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