Question

car a is traveling to the right at a constant velocity $v_a$. at time $t = 0$, it passes car b, which is at rest. at the same time ($t = 0$), car b begins to accelerate with a constant acceleration of magnitude $a_b$, as shown in figure 1. car b has a velocity of $v_b$ when it reaches the same position as car a at time $t = t_f$, as shown in figure 2. which of the following, if any, is an expression for the time it takes for car b to catch up to car a? a $\frac{v_a}{a_b}$ b $\frac{2v_a}{a_b}$ c $\frac{2v_a^2}{a_b}$ d it cannot be determined without knowing the distance traveled by the cars.

Explanation:

Step1: Write displacement formula for Car A

Car A has a constant - velocity motion. The displacement formula for an object with constant velocity is $x = v_0t$. For Car A, $x_A=v_At$, where $v_A$ is the velocity of Car A and $t$ is the time.

Step2: Write displacement formula for Car B

Car B has a constant - acceleration motion. The displacement formula for an object with constant acceleration starting from rest ($v_{B0} = 0$) is $x=v_{0}t+\frac{1}{2}at^{2}$. Since $v_{B0} = 0$, for Car B, $x_B=\frac{1}{2}a_Bt^{2}$, where $a_B$ is the acceleration of Car B.

Step3: Set displacements equal

When Car B catches up with Car A, $x_A=x_B$. So, $v_At=\frac{1}{2}a_Bt^{2}$.

Step4: Solve for time $t$

First, move all terms to one side of the equation: $\frac{1}{2}a_Bt^{2}-v_At = 0$. Factor out a $t$: $t(\frac{1}{2}a_Bt - v_A)=0$. One solution is $t = 0$ (corresponds to the initial time when Car A just passes Car B). To find the non - zero solution, set $\frac{1}{2}a_Bt - v_A=0$. Then, $\frac{1}{2}a_Bt=v_A$, and $t=\frac{2v_A}{a_B}$.

Answer:

B. $\frac{2v_A}{a_B}$