Source: Principles of Physics
Problem Sets:
Problem
Car A starts at position $x_{A0}$ with a constant velocity $v_A$. Car B starts at the origin with an initial velocity $v_{B0}$ and a constant negative acceleration $a_B$. Assume $v_{B0} > v_A > 0$ and $x_{A0} > 0$.
- What must $a_B$ be such that the cars are at the same position at time $t_1 > 0$?
- For this acceleration, what is the other time $t_2$ when the cars are at the same position?
- Find the critical magnitude of acceleration, $A_{crit} = |a_B|_{crit}$, for which the cars meet only once.
- For $A = |a_B|$, how many times do the cars meet if $A > A_{crit}$ or if $A < A_{crit}$?
[Q1] $a_B = \frac{2(x_{A0} - (v_{B0}-v_A)t_1)}{t_1^2}$ [Q2] $t_2 = \frac{-2x_{A0}}{a_B t_1}$ [Q3] $A_{crit} = \frac{(v_{B0}-v_A)^2}{2x_{A0}}$ [Q4] If $A > A_{crit}$: 0 times. If $A < A_{crit}$: 2 times.
The positions of the cars are given by:
$x_A(t) = x_{A0} + v_A t$ $x_B(t) = v_{B0}t + \frac{1}{2}a_B t^2$The cars are at the same position when $x_A(t) = x_B(t)$. This gives a quadratic equation for time $t$:
$$x_{A0} + v_A t = v_{B0}t + \frac{1}{2}a_B t^2$$ $$\frac{1}{2}a_B t^2 + (v_{B0} - v_A)t - x_{A0} = 0$$[Q1] For the cars to meet at time $t_1$, it must be a root of the equation. We substitute $t = t_1$ and solve for $a_B$:
$$\frac{1}{2}a_B t_1^2 + (v_{B0} - v_A)t_1 - x_{A0} = 0 \implies a_B = \frac{2}{t_1^2} [x_{A0} - (v_{B0} - v_A)t_1]$$[Q2] Let the two roots of the quadratic equation be $t_1$ and $t_2$. By Vieta's formulas, the product of the roots is $t_1 t_2 = c/a_q = (-x_{A0}) / (\frac{1}{2}a_B)$.
$$t_2 = \frac{-2x_{A0}}{a_B t_1}$$[Q3] The cars meet only once if the quadratic equation has exactly one positive real root. Let $A = |a_B| = -a_B$. The equation becomes $\frac{1}{2}A t^2 - (v_{B0} - v_A)t + x_{A0} = 0$. This occurs when the discriminant $\Delta = 0$.
$$\Delta = (-(v_{B0}-v_A))^2 - 4(\frac{1}{2}A)(x_{A0}) = (v_{B0}-v_A)^2 - 2Ax_{A0}$$Setting $\Delta = 0$ gives the critical acceleration $A_{crit}$:
$$A_{crit} = \frac{(v_{B0}-v_A)^2}{2x_{A0}}$$[Q4] The number of real roots is determined by the sign of $\Delta$. If $A > A_{crit}$, then $\Delta < 0$, so there are no real roots. The cars never meet. If $A < A_{crit}$, then $\Delta > 0$, so there are two distinct real roots. Since $v_{B0} > v_A$ and $x_{A0} > 0$, both roots are positive. The cars meet twice.