1. ## Rectilinear Motion Help

I am just learning how to do this for my calculus class so forgive me for being so ignorant, but I am trying.

For the problems I am doing, these two formulas are being used:

v(t)= v(sub zero) + at
s(t) = s(sub zero) + v(sub zero)t + .5at^2

I'm having a lot of trouble with applying these. For example, the current problem I am on says:

"Spotting a police car, you hit the brakes on your new Porsche to reduce your speed from 90mi/h to 60mi/h at a constant rate over a distance of 200 feet. (88ft/s=60mi/h)

Find acceleration in ft/s^2"

I need to understand which formula is appropriate for various situations. I realize:
initial velocity= 132ft/s
final velocity= 88ft/s
initial displacement=0ft
final displacement= 200ft

I have no idea how to find time to solve for acceleration, so both formulas seem useless to me.

If someone could explain these formulas and help me on this current problem, I would really appreciate it. I've been on this same problem for 3 hours and have gotten nowhere.

2. Originally Posted by Simon777
For the problems I am doing, these two formulas are being used:

v(t)= v(sub zero) + at
s(t) = s(sub zero) + v(sub zero)t + .5at^2

I'm having a lot of trouble with applying these. For example, the current problem I am on says:

"Spotting a police car, you hit the brakes on your new Porsche to reduce your speed from 90mi/h to 60mi/h at a constant rate over a distance of 200 feet. (88ft/s=60mi/h)

Find acceleration in ft/s^2"

I need to understand which formula is appropriate for various situations. I realize:
initial velocity= 132ft/s
final velocity= 88ft/s
initial displacement=0ft
final displacement= 200ft

I have no idea how to find time to solve for acceleration, so both formulas seem useless to me.
You need to put the given information into both those formulas.

For the formula $v(t)= v_0 + at$, you have $v(t) = 88$ and $v_0 = 132$, which tells you that $at = -44$ (it's negative because the car is decelerating, so $a$ is negative).

For the formula $s(t) = s_0 + v_0t + \tfrac12at^2$, you have $s(t) = 200$, $s_0=0$ and $v_0=132$. Plug those into the formula and you get $200 = 132t + \tfrac12(at)t$. I have written the last term $\tfrac12at^2$ as $\tfrac12(at)t$, because you can now substitute in the value for $at$ that you got from the previous equation. You then have a simple equation for $t$, and once you know $t$ you can put it into the formula $at = -44$ to find $a$.

3. Originally Posted by Opalg
You need to put the given information into both those formulas.

For the formula $v(t)= v_0 + at$, you have $v(t) = 88$ and $v_0 = 132$, which tells you that $at = -44$ (it's negative because the car is decelerating, so $a$ is negative).

For the formula $s(t) = s_0 + v_0t + \tfrac12at^2$, you have $s(t) = 200$, $s_0=0$ and $v_0=132$. Plug those into the formula and you get $200 = 132t + \tfrac12(at)t$. I have written the last term $\tfrac12at^2$ as $\tfrac12(at)t$, because you can now substitute in the value for $at$ that you got from the previous equation. You then have a simple equation for $t$, and once you know $t$ you can put it into the formula $at = -44$ to find $a$.
I didn't even think about substituting, thank you.