First slide

Rectilinear Motion

Question

A fun drive in an amusement park runs between two spots that are 2.0 km apart. For safety reasons the acceleration of the drive is limited to $\pm$ 4.0 m/s², and the jerk, or rate of change of acceleration, is limited to $\pm$ 1.0 m/s². The drive has a maximum speed of 144 km/h. If the shortest time taken by the drive to travel between the spots is n², The value of 'n' is ______.

Difficult

Solution

$\begin{array}{l} \frac{d a}{d t} = 1 \Rightarrow a = t \\ Let at t = t_{1}, a = 4 \Rightarrow t_{1} = 4 s \end{array}$
$\begin{array}{l} \frac{d v}{d t} = t \Rightarrow v = \frac{t^{2}}{2} \\ At t = t_{1}, v_{1} = \frac{4^{2}}{2} = 8 m / s \\ \frac{d x}{d t} = \frac{t^{2}}{2} \Rightarrow x = \frac{t^{3}}{6} \end{array}$
Let at
$\begin{array}{l} t = t_{2}, v_{2} = 144 km / h = 40 m / s \\ v_{2} = v_{1} + a t_{2} \Rightarrow 40 = 8 + 4 t_{2} \\ \Rightarrow t_{2} = 8 s x_{1} = \frac{4^{3}}{6} = \frac{32}{8} m \\ x_{2} = 8 \times 8 + \frac{1}{2} \times 4 (8)^{2} = 192 km \end{array}$
Distance for which it runs with uniform velocity:
$\begin{array}{l} S = 2000 - 2 (x_{1} + x_{2}) t_{3} = \frac{s}{40} \\ Total time: T = t_{1} + t_{2} + t_{3} + t_{2} + t_{1} = 64 s \end{array}$

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