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- Homework Statement
- Two cars start at rest at the same point and travel in the same direction. The first car leaves at ##t_1 = 0## with an acceleration of ##a_1 = 3 m/s^2##. The second car leaves at ##t_2 = 6## with an acceleration of ##a_2 = 5 m/s^2##. At what time does the second car overtake the first?

- Relevant Equations
- $$v_x = v_{0x} + \int_0^t a_x \, dt$$

$$x = x_0 + \int_0^t v_x \, dt$$

So this isn't actually a homework question, it's a question that was on the test I just took and I cannot figure out what I'm doing wrong. I've tried solving it three or four different ways, double checking everything, and I keep getting ##15 + 3\sqrt {15}## or ~26.6, which was not one of the answer choices.

Starting by finding ##x_1(t),## I know initial position and velocity are both zero so I can find ##x_1(t)## by integrating ##a_1 = 3 m/s^2## twice. $$v_1(t) = \int_0^t 3 \, dt = 3t$$

$$x_1(t) = \int_0^t 3t \, dt = \frac 3 2 t^2$$

To find ##x_2(t),## I did the same thing but I had to find initial position and velocity as well.

$$v_2(t) = v_0 + \int_0^t 5 \, dt$$

$$v_2(t) = v_0 + 5t$$

I know ##v_2(6) = 0,## so

$$0 = v_0 + 5(6)$$

$$v_0 = -30$$

Therefore, $$v_2(t) = 5t - 30$$

Then, I did the same thing again to find ##x_2(t)##

$$x_2(t) = x_0 + \int_0^t 5t - 30\, dt$$

$$x_2(t) = x_0 + \frac 5 2 t^2 - 30 t$$

$$ 0 = x_0 + \frac 5 2 (6)^2 - 30 (6)$$

$$0 = x_0 + 90 - 180$$

$$x_0 = 90$$

$$x_2(t) = \frac 5 2 t^2 - 30t + 90$$

Finally, to find the final answer, I set them equal to each other and solved for t

$$\frac 3 2 t^2 = \frac 5 2 t^2 - 30t + 90$$

$$0 = t^2 - 30t + 90$$

$$t = \frac {30 \pm \sqrt{30^2 -4(90)}} {2},$$

$$t = 15 \pm 3 \sqrt 15$$

$$t = 3.38, t = 26.6$$

And I threw out ##t = 3.38## because it's less than 6.

Starting by finding ##x_1(t),## I know initial position and velocity are both zero so I can find ##x_1(t)## by integrating ##a_1 = 3 m/s^2## twice. $$v_1(t) = \int_0^t 3 \, dt = 3t$$

$$x_1(t) = \int_0^t 3t \, dt = \frac 3 2 t^2$$

To find ##x_2(t),## I did the same thing but I had to find initial position and velocity as well.

$$v_2(t) = v_0 + \int_0^t 5 \, dt$$

$$v_2(t) = v_0 + 5t$$

I know ##v_2(6) = 0,## so

$$0 = v_0 + 5(6)$$

$$v_0 = -30$$

Therefore, $$v_2(t) = 5t - 30$$

Then, I did the same thing again to find ##x_2(t)##

$$x_2(t) = x_0 + \int_0^t 5t - 30\, dt$$

$$x_2(t) = x_0 + \frac 5 2 t^2 - 30 t$$

$$ 0 = x_0 + \frac 5 2 (6)^2 - 30 (6)$$

$$0 = x_0 + 90 - 180$$

$$x_0 = 90$$

$$x_2(t) = \frac 5 2 t^2 - 30t + 90$$

Finally, to find the final answer, I set them equal to each other and solved for t

$$\frac 3 2 t^2 = \frac 5 2 t^2 - 30t + 90$$

$$0 = t^2 - 30t + 90$$

$$t = \frac {30 \pm \sqrt{30^2 -4(90)}} {2},$$

$$t = 15 \pm 3 \sqrt 15$$

$$t = 3.38, t = 26.6$$

And I threw out ##t = 3.38## because it's less than 6.