Question

A car starts from rest and accelerates uniformly at 3.0 m/s^2. A second car starts from rest 6.0s later at the same point and accelerates uniformly at 5.0m/s^2. How long does it take the second car to overtake the first?

Answer 1

A. 19s
B. 12s
C. 24s
D. 21s

Answer 2

First, you need the equations of motion of both cars. Let the first car be A, and the second B,
\[x_A=x_{0A}+v_{0A}t+a_At^2/2\]\[x_B=x_{0B}+v_{0B}t+a_Bt^2/2\]
Apply the initial conditions,
\[x_A=3t^2/2\]\[x_B=5(t-6)^2/2\]
Then, you need the instant when both cars share their position,
\[x_A=x_B\Rightarrow 3t^2/2=5(t-6)^2/2\Rightarrow 3t^2=5(t^2+36-12t)\]
\[2t^2-60t+180=0\Rightarrow t^2-30t+90=0\Rightarrow t\approx26.6\ s\]

Answer 3

Well, this time is measured from t=0, taking the 6 s you have,
\[26.6-6=20.6\approx21\ s\]

Answer 4

Maybe this will help explain the process as well. Start with this kinematic equation for constant acceleration:\[\large \Delta x = V_it+\frac{1}{2}at^2\]The initial velocity for both cars will be 0m/s so this simplifies to:\[\large \Delta x = \frac{1}{2}at^2\]When car B overtakes car A delta X will be the same, so:\[\large \frac{1}{2}a(t+6)^2=\frac{1}{2}at^2\]Which simplifies to:\[\large a(t+6)^2=at^2\]Now plug in the values and solve for t:\[\large (3.0m/s^2)(t+6)^2=(5.0m/s^2)t^2\]You should end up with two values...discard the negative one and you end up with 20.6s which obviously rounds to 21s.