Stopping Distance (frictionless) Assuming I have a body travelling in space at a rate of $1000~\text{m/s}$. Let's also assume my maximum deceleration speed is $10~\text{m/s}^2$. How can I calculate the minimum stopping distance of the body?
All the formulas I can find seem to require either time or distance, but not one or the other.
 A: The formula you want is
$$v_f^2 = v_i^2 + 2a(x_f - x_i)$$
It's one of the basic kinematic formulas taught in high school (or even middle school) physics classes.
A: If the speed is $1000 m/s$ and the deceleration is $10 m/s^2$, it will take $100 s$ to stop.  The average speed in that time is $500 m/s$, so the distance traveled is 
$$500m/s*100s = 5*10^4m$$
Working through the same logic with an initial speed $v$ and a deceleration $a$, the final distance $d$ traveled before stopping is
$$d = v_{avg}*t = (v/2)*(v/a) = \frac{v^2}{2a}$$
This formula becomes more interesting when you learn a bit more physics because it's simple example of the work-energy theorem.
A: Another equation of motion problem,very easy. 
$$v^2 = u^2 + 2.a.s$$ where $$v = \textbf{final velocity} $$ , $$u = \textbf{initial velocity}$$ , $$a=\textbf{ acceleration or in this case negative acceleration}$$ , $$s = displacement$$ . This equation is time independent. Now, to find $s$ , put the values: $$s = \frac{-1000^2}{2.(-10)} \implies s = 50000~\text{m}$$ . Simple,right?
A: Double integrate the acceleration with respect to time.  This first integral will give you the velocity graph based on the acceleration.  With the first integral, add a constant and find the constant by equating the first indefinite integral to the initial velocity at t=0.  This initial value will correct the deceleration for motion at time t=0.  The constant will allow for the second integration.  Now find at what time the first integral will equal 0 (the time when velocity equals 0). Integrate a second time with respect to time and substitute in the time figure of 0 and the final time and solve.  This will give you the full distance to stop.
integrate(a) dt = at + c; integrate(at + c) dt = ((at^2)/(2)) + ct; Sub in final time and solve, then sub in initial time (0) and subtract initial from final.
A: *

*When brakes are applied to a moving vehicle, the distance it travels before stopping is called the stopping distance.

*It is an important factor for road safety and depends on the initial velocity ($v_0$) and the braking capacity, or the deceleration, $–a$ that is caused by the braking.

*Let the distance travelled by the vehicle before it stops be $d_s$. Then, using equation of motion $v^2 = {v_{o}}^2 + 2 ax$, and noting that $v = 0$, we
have the stopping distance
$$d_s = \frac{{-v_{0}}^2}{2a} $$

*Thus, the stopping distance is proportional to the square of the initial velocity. Substituting ${v_{o}}^2 = 1000~\text{m/s}$ and $a = -10~\text{m/s}^2$ in the above expression, we get $d_s = 50000~\text{m}$.

