find distance traveled of object that is slowing down I apologize in advance because I feel like this problem is fairly simple but I can't seem to figure out what the formula would be.
Essentially, if I had an object that were traveling 60cm/second, but had an inverse force applied to it of -60cm/second, how would I calculate how much distance it would travel in 1 second before it stopping? (Ignore factors such as wind resistance, gravity, etc. Only account for velocity of the object, time, and the inverse force applied to it).
 A: Standard uniform deceleration problem. You did mean decelerative ( negative acceleration ) $ 60 \dfrac { cm}{sec^{2}}, $
Dynamic relations are
$$ v = u + a t , s = u t + a t^2/2. $$
The latter relation is useful, plug in $ u=60, a = -60, t=1. $
A: Your object has a speed of $(60-60t)$ cm/sec. At what time is the speed zero?  Now integrate the velocity from $0$ to this time to find the displacement.
A: $$distance = \int_0^1 v(t)dt=\int_0^1 60(1-t) dt=\left[60t-30t^2\right]_0^1=30cm$$
A: If its acceleration of $-60cm/second^2$ then we have equations of motion Which calculates distance of particle. $S_n=u+\frac{a}{2}(2n-1)$ substituing $n=1$ we get it as $60-\frac{60}{2}(2-1)=30$. $'-'$ indicates deacceleration. Hope its clear.
A: One way to solve this problem is to look at energy and work. The object starts off with a kinetic energy of $\frac12mv_0^2$. To bring it to a complete stop, that much work must be done to it. For a constant force directed along the line of motion, work is equal to force $\times$ distance, so we have $Fs=mas=-\frac12mv_0^2$, so $s=-\frac{v_0^2}{2a}$. (The minus sign is because the change in kinetic energy is negative: $\Delta E_k=0-\frac12mv_0^2$.) Plugging in your values (and assuming you really meant an acceleration of $-60\;\text{cm}/\text{sec}^2$), we get $$s = -\frac{(60\;\text{cm}/\text{sec})^2}{2\cdot (-60\;\text{cm}/\text{sec}^2)} = 30\;\text{cm}.$$ 
Even though for this problem it’s easy enough to compute the time it will take to stop the object and then integrate over that time interval, this method requires no integrals at all.
