# Finding deceleration and initial speed, given time and distance

(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.

• Are we assuming deceleration is also constant in second case? – user160738 Jun 20 '17 at 3:49
• yes the deceleration is constant. Also both car stops so their final velocity is 0m/s. – Kong Nyian Sii Jun 20 '17 at 3:58
• 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. – user160738 Jun 20 '17 at 4:09

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 $100=4v_0, v_0=25, a=-\frac {25}8$