# Thread: Antiderivative and acceleration question

1. ## Antiderivative and acceleration question

At the instant the traffic light turns green, an automobile that has been waiting at an intersection starts moving forward with a constant acceleration of 6 feet per second squared. At the same instant, a truck traveling with a constant velocity of 30 feet per second overtakes and passes the car. How far beyond its starting point will the automobile overtake the truck?

I'm so lost, any help would be appreciated!

EDIT: Sorry this is my first time posting, and I did not realise I should show work. I also do not know how to write in math text. All I know is that the integral of acceleration in equal to velocity, and the integral of velocity is equal to position. I still do not know how to approach this problem, however.

2. The equation you need here is the can be derived in two lines from the fact that the acceleration is the double derivative of the displacement. And for both vehicles the acceleration is constant (either 6 or 0 ft/s).

$\frac{d^2s}{dt^2} = a.$

Integrating

$\frac{ds}{dt} = v_0 + at,$

where $v_0$ is the velocity of the vehicle at the time $t=0$. Integrating again gives

$s = s_0 + v_0 t + \frac{1}{2} at^2,$

where $s_0$ is the displacement of the vehicle at the time $t=0$. In this problem they are both at the traffic light which we will take as $s=0$ and so $s_0$ can be ignored in this solution.

So the displacement of the car $s_c$ is easy as in this case $v_0 = 0$ and $a = 6$, so

$s_c = 3t^2.$

And in the case of the truck $s_t$, $v_0 = 30$ and $a = 0$

$s_t = 30t.$

Now we want the time $t_1$ where the car overtakes the truck. IE displacements are equal therefor $s_c = s_t$

$3t_1^2 = 30t_1$

$t_1 = 10$

And finally to find what that distance is $s_1$ you can substitute $t=t_1$ into either the $s_c$ or the $s_t$ equation as they will give the same answer.

$s_1 = 30 t_1 = 300.$

So the car will overtake the truck 300 ft from the traffic light.

To make sure you have understood this, try working out how fast the car will be travelling at the point that it overtakes the truck.

Hope this helps.