Get acceleration from distance and speed

Question

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

---

Thanks in advance, Bruno

Answer 1

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 $a$ = $\frac{-(v_1)^2}{2d}$.
$a$ = $\frac{-(2.77)^2}{2(2.35)}$ $\approx -1.63253 \frac{m}{s^2}$

Answer 2

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

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