My basic problem is this:

In a frictionless environment, I have an object traveling at a known speed that has a known maximum deceleration rate and a known distance to move before it must stop. I want to calculate how much acceleration I can apply prior to "hitting the brakes" over the time period of 1/60th a second. I have the data below to use, though much of it is variable until the time the calculation is done:

  • The available deceleration in $\mathrm{m}\,\mathrm{s}^{-2}$, known at the time of the equation
  • Current velocity $\mathrm{m}\,\mathrm{s}^{-1}$, known at the time of the equation
  • Distance before the object must come to a complete stop in m, known at the time of the equation
  • Period of acceleration - 1/60 s

I had the formula for maximum speed I could go to reach a speed of 0 in a given distance with a given acceleration as the below:

velocity = (distance * acceleration * 2)$^{-2}$

I found the formula for distance travelled under constant acceleration as:

distance = initial velocity * time + acceleration / 2 * (time) $^ 2$

Initially, I had thought to do the below:

Acceleration = Max velocity now - Max velocity then (using the distance formula to subtract from the initial distance in it) / time period

However after going through and trying to solve for acceleration (I ended up with a quadratic equation which doesn't seem right), I realized I messed up someplace as this only helps me if I am moving at the current maximum speed and will give me how quickly I must slow down, which isn't very valuable to me and could be found much easier if it were.

I'm betting I'm overthinking and I'm missing a simple way to do this, but I can't for the life of me think of it. Any insight at all would be greatly appreciated.


If the braking distance is $s$ and the maximum deceleration is $a$ then the maximum speed from which stopping is possible within this distance is $v=\sqrt{2as}$. This comes from the kinematic equation $v^2=u^2+2as$.

The maximum acceleration from velocity $u$ up to velocity $v$ over a time of $t=\frac{1}{60}s$ immediately prior to braking, is $A=\frac{v-u}{t}=\frac{\sqrt{2as}-u}{t}$. This comes from the kinematic equation $v=u+at$. Any greater value of $A$ will result in the speed exceeding $v=\sqrt{2as}$ when the brakes are applied, so that you are not able to stop within the given distance.

  • $\begingroup$ Yeah I way over complicated this for myself. Thank you for your quick and concise answer! $\endgroup$ – Vivicorp Feb 17 '17 at 2:01

If you have a certain acceleration $a$ and have to come to rest at a distance $s$, your velocity should be $v=\sqrt{2as}$. Any velocity greater than that then, your stopping distance is greater than $s$ and vice versa.

So, max. velocity you can have is $\sqrt{2\times\text{max}\, a\times s}$


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