Answer to Question #94139 in Mechanics | Relativity for stefanie

Question

At t = 0, one toy car is set rolling on a straight track with initial position 13.5 cm, initial velocity -4.4 cm/s, and constant acceleration 2.80 cm/s2. At the same moment, another toy car is set rolling on an adjacent track with initial position 9.5 cm, initial velocity 5.00 cm/s, and constant zero acceleration.

At what time(s), if any, do the cars pass each other?

What are their locations at that time?

Answer 1

In the below notation the location of the 1st car will be x and the location of the 2nd car will be y (please note these axis are parallel, not perpendicular, in this notation).

$x_0 = 13.5 cm; v_0 = -4.4\frac{cm}{s}; a_0 = 2.80 \frac{cm}{s^2}; \\ y_0 = 9.5 cm; u_0 = 5.00\frac{cm}{s}; b_0 = 0; \\ x(t) = x_0 + v_0t + \frac{a_0t^2}{2}; \\ y(t) = y_0 + u_0t + \frac{b_0t^2}{2} = y_0 + u_0t; \\$

Let T be the time when two cars meet.

$x(T) = y(T); \\ \frac{a_0T^2}{2} + v_0T + x_0 = u_0T + y_0; \\ \frac{1}{2}a_0T^2 + (v_0-u_0)T + (x_0-y_0) = 0; \\ T_{1,2} = \frac{(u_0-v_0) \pm \sqrt{(u_0-v_0)^2 - 2a_0(x_0-y_0)}}{a_0};$

We will make the calculation with the numerical values only, dropping the units, to simplify the notation. It is easy to check that the dimension of the result will be correct (units are seconds).

$T_{1,2} = \frac{(5+4.4) \pm \sqrt{(5+4.4)^2 - 2\cdot2.8\cdot(13.5-9.5)}}{2.8} \approx 0.46\text{ or }6.26\text{ [s]}. \\$

The corresponding locations we will get by introducing the numerical values of T into the equation for y(t) (as simpler one):

$x(T_1) = y(T_1) \approx 9.5 + 5\cdot0.46 \approx 11.78\text{ [cm]} ; \\ x(T_2) = y(T_2) \approx 9.5 + 5\cdot6.26 \approx 40.79\text{ [cm]} .$