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(I'm sorry if this is a bad way to frame a math question)

Let's say a car is moving at 40m/s and a brake is applied, the car decelerates at 10m/s thus stops after 4 seconds, traveled 100m after the brake is applied.

Now a different car is moving at unknown speed, a brake with different deceleration is applied and the car stops after 8 seconds, but also traveled 100m.

Is it possible to find out what is the initial speed AND the deceleration of the second car?

Example Picture

I have been trying to find if there's a formula to this, but i can only find either the formula include time or distance, but not both.

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  • $\begingroup$ Are we assuming deceleration is also constant in second case? $\endgroup$
    – user160738
    Jun 20, 2017 at 3:49
  • $\begingroup$ yes the deceleration is constant. Also both car stops so their final velocity is 0m/s. $\endgroup$
    – Kong Nyian Sii
    Jun 20, 2017 at 3:58
  • $\begingroup$ If $v_0$ was initial velocity, velocity of the car at time $t$ after applying brake would be $v=(1-0.125t)v_0$. Integrating this with respect to $t$, you have $x(t)=v_0(t-\frac{t^2}{16})$. When the car stopped it travelled $100$m, so $x(8)=100$. This gives you both $v_0$ and deceleration. $\endgroup$
    – user160738
    Jun 20, 2017 at 4:09

2 Answers 2

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You can use the usual $s=x_0+v_0t+\frac 12at^2, v=v_0+at$ formulas. We are given $x_0=0$ because we measure from there. We are given $v(8)=0$, so $a=-\frac {v_0}8$. We are also told $s(8)=100$ You have two equations in two unknowns, $a, v_0$.

Added: plugging in to $s(8)=100$ we get $100=8v_0+\frac 12 (-\frac {v_0}8)8^2$ or $100=4v_0, v_0=25, a=-\frac {25}8$

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  • $\begingroup$ As a layman this took me entire day to find figure out what this actually is. It turns out the example that I have provided is not actually practical, as it calculates the outcome by several ticks rather than precision. It doesn't seem possible to calculate such things, at least to my knowledge. But I really appreciate all of you for pointing me to the right direction. Now I understand the problem even better. $\endgroup$
    – Kong Nyian Sii
    Jun 20, 2017 at 17:09
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I mean you could use the kinematic equations

Vf^2 = Vi^2 + 2A(Xf-Xi) Xf - Xi =Vit + .5At^2 and then rearrange to get Vf^2 = (( Xf - Xi - .5At^2)/t)^2+ 2A(Xf-Xi) We know that Vf = 0 Soo 0 = (( 100 - 32A)/8)^2 + 200A No you can rearrange and solve for A

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