Get acceleration from distance and speed [closed]

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...

Given:

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

Searching:

• deccelaration

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

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• Write down the relevant equations. – David G. Stork Sep 24 '18 at 16:47
• See Equation 4. – Math Lover Sep 24 '18 at 16:49
• @David G. Stork edited OP – Bruno Sep 24 '18 at 17:01
• @Math Lover There are some interesting formulas there, but not the one I am looking for. – Bruno Sep 24 '18 at 17:03
• @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. – 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 $$a$$ = $$\frac{-(v_1)^2}{2d}$$.
$$a$$ = $$\frac{-(2.77)^2}{2(2.35)}$$ $$\approx -1.63253 \frac{m}{s^2}$$

• 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? – 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)

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

or

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