I want to know how fast an object with the speed s1 (2.77m/s) would have to deccelarate to reach a speed s2 (0m/s) within a certain distance d (2.35m). With a formula for that I could go on to calculate the time needed, the force applied and so on...


  • Initial speed (2.77m/s)
  • resulting speed (0m/s)
  • distance to deccelarate (2.35m)


  • deccelaration

Thanks in advance, Bruno


closed as off-topic by Adrian Keister, Math Lover, Delta-u, let's have a breakdown, Theoretical Economist Sep 24 '18 at 19:27

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  • $\begingroup$ Write down the relevant equations. $\endgroup$ – David G. Stork Sep 24 '18 at 16:47
  • 1
    $\begingroup$ See Equation 4. $\endgroup$ – Math Lover Sep 24 '18 at 16:49
  • $\begingroup$ @David G. Stork edited OP $\endgroup$ – Bruno Sep 24 '18 at 17:01
  • $\begingroup$ @Math Lover There are some interesting formulas there, but not the one I am looking for. $\endgroup$ – Bruno Sep 24 '18 at 17:03
  • $\begingroup$ @Bruno Eq. 4 in the said link should solve your problem. You might look at some 'simple' forms of these equations beneath the first set of equations. $\endgroup$ – Math Lover Sep 24 '18 at 17:06

We have the following formula, which can be rearranged easily.
$(v_2)^2 - (v_1)^2$ = $2ad$ $\Rightarrow$ $a$ = $\frac{(v_2)^2-(v_1)^2}{2d}$.
Since our deceleration leads to a final velocity of $0 \frac{m}{s}$, we can say $a$ = $\frac{-(v_1)^2}{2d}$.
$a$ = $\frac{-(2.77)^2}{2(2.35)}$ $\approx -1.63253 \frac{m}{s^2}$

  • $\begingroup$ How would I test if that result is correct? Can I just create a function f(x)=-1.63253x^2+2.77 and look at the interval from x=0 to y=0? $\endgroup$ – Bruno Sep 24 '18 at 17:20

It is an easy kinematics/physics problem. Your object movement is described by movement equations.

$s2=s1-a*t (1)$

$d = s1*t - \frac{a*t^2}{2} (2)$

where s1,s2 -velocity, t - time, a - deceleration, d-distance.

Because $S2=0$, so (1) can be rewritten

$s1=a*t => t=\frac{s1}{a}(1')$

put (1') to (2)

$d = \frac{s1^2}{a} - \frac{s1^2}{2*a} = \frac{s1^2}{2*a}(3)$


$a = \frac{s1^2}{2*d} = \frac{2.77^2}{2*2.35} = 1.6325 \frac{m}{s^2}$


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