Question

A ramp is 350m long. A bar starting at rest accelerates down the ramp at 1.2 m/s^2. 10s later a bowling ball is pushed down at 5m/s and accelerates 1.5m/s^2. What is the closest approach on the RAMP? Please help I have been trying forever.

Answer 1

Question 35633

Let the ox axis of the coordinate system be parallel to the inclined plane of the ramp. Position of bar as a function of time is $x_{b}(t) = \frac{a_{b}t^{2}}{2}$ , where $a_{b} = 1.2\frac{m}{s^{2}}$ is the given acceleration. Since bowling ball is pushed down $t_1 = 10s$ later, its position as a function of time is $x_{bb}(t) = v_0(t - t_1) + \frac{a_{bb}(t - t_1)^2}{2}$ , where $v_{0} = 5\frac{m}{s^{2}}$ is the initial velocity of the bowling ball and $a_{bb} = 1.5\frac{m}{s^{2}}$ is the acceleration of the bowling ball. When bar and bowling ball meet each other, $x_{b}(t) = x_{bb}(t)$ .

This yields an equation for time: $v_{0}(t - t_{1}) + a_{bb}\frac{t^{2}}{2} - a_{bb}t t_{1} + a_{bb}\frac{t_{1}^{2}}{2} = \frac{a_{b}t^{2}}{2}$ , which might be rewritten as $\frac{t^2}{2} (a_{bb} - a_b) + t(v_0 - a_{bb}t_1) - [v_0t_1 - a_{bb}\frac{t_1^2}{2}] = 0$ . This is simple quadratic equation and has solutions $t_1 = 2.6s; t_2 = 64.07s$ . One should choose solution $t_1$ (for closest approach).

Finally, plugging in $t_1$ into $x_b(t)$ or $x_{bb}(t)$ finds the distance $L = x_b(t)|_{t=2.6s} = 4.06m$ .