Two Bikes Syncing Up

Classical Mechanics Level 4

Two bicycles (and riders) are on a straight, one-dimensional path. $(x_1,v_1,a_1)$ represent the position, velocity, and acceleration of Bike 1, and $(x_2,v_2,a_2)$ represent the position, velocity, and acceleration of Bike 2.

The following conditions apply:

At time $t = 0$ :

$x_1 = 0 \\ v_1 = 0 \\ x_2 = 10 \\ v_2 = 5$

Acceleration parameters:

$a_1 = \alpha \hspace{1cm} \text{for} \,\,\, 0 \leq t < T \\ a_1 = -\alpha \hspace{1cm} \text{for} \,\,\, T \leq t \leq 20 \\ a_2 = 0 \hspace{1cm} \text{for} \,\,\, 0 \leq t \leq 20$

At time $t = 20$ :

$x_1 = x_2 \\ v_1 = v_2$

Parameters $\alpha$ and $T$ are deliberately withheld. Determine the value of $\alpha$ .

The answer is 0.6905.

1 solution

Miles Koumouris
Dec 12, 2017

Since $v_1=\left\{ \begin{array}{lll} \alpha t & \hspace{1cm} & 0 \leq t < T \\-\alpha (t-T)+\alpha T & & T \leq t < 20, \end{array}\right.$ it is evident that $T=\dfrac{20\alpha +5}{2\alpha }.$ Now $\int_0^T\alpha t\; dt\; +\; \int_T^{20}-\alpha (t-T)+\alpha T\; dt \; = \; 110.$ Note that $\alpha >0$ since the displacement is positive, and hence $\begin{aligned} 400\alpha^2-240\alpha -25&=0\\ \Longrightarrow \alpha &=\dfrac{\sqrt{61}+6}{20}\approx \boxed{0.6905}. \end{aligned}$

Two Bikes Syncing Up

The answer is 0.6905.

1 solution

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