1
$\begingroup$

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

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, Math Lover, Delta-u, let's have a breakdown, Theoretical Economist
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.