My brain hurts and I'd really appreciate some help with this formula. I am currently at 0m travelling 50m/s. I want to decelerate and come to a stop at 10m in exactly 0.5 seconds. What should be my acceleration to achieve this.
No mechanics formula I know of lets me specify a start speed, end speed, distance, and time and gives me an acceleration. Any help is appreciated.
You know that $$x=x_0+v_0t+\frac{1}{2}at^2$$ $$v=v_0+at$$
with $\,x_0=0,\,v_0=50,\,x=100,v=0\,t=0.5$.
So, $$0=50+0.5a\Rightarrow a=-100\,[\text{m/s}^2]$$
But $$0+50\cdot0.5-\frac{1}{2}(-100)0.5^2=12.5\,[\text{m}]$$
And that's not even near of what you want.
So, I'm inclined to use a piecewise acceleration function (accelerate with $a_1$ until you hit $X$ meters, and then deccelerate with $a_2$ to exactly stop at the 100m mark).
Since other users showed that it can't be constant. Let's check the next simplest type: linear.
So assume
$$a(t) = mt+b$$
We know that $$v(t)-v(0)=\int_0^ta(t)\,\mathrm dt$$ and $$x(t)-x(0)=\int_0^t v(t)\,\mathrm dt$$
Starting with the first equation, we know that
$$0-50=\left[\frac{1}{2}mt^2+bt\right]_0^{0.5}$$
which simplifies to $$\frac{m}{8}+\frac{b}{2}=-50$$
Next we know that
$$v(t)=v(0)+\int_0^ta(t)\,\mathrm dt=v(0)+\left[\frac{1}{2}mt^2+bt\right]_0^t=50+\frac{1}{2}mt^2+bt$$
$$x(0.5)-x(0)=100=\int_0^{0.5}50+\frac{1}{2}mt^2+bt\,\mathrm dt$$
which simplifies to $$\frac{m}{48}+\frac{b}{8}=75$$
The intersection of these lines is at $m=-8400,b=2000$
So $$\boxed{a(t)=\left(-8400\frac{\mathrm{m}}{\mathrm{s}^3}\right)t+2000\frac{\mathrm{m}}{\mathrm{s}^2}}$$
Let's check our math.
Assume $a(t)=-8400t+2000$.
$$v(t)=\int_0^ta(t)\,\mathrm dt=\int_0^t-8400t+2000\,\mathrm dt=-4200t^2+2000t+c$$
Well $v(0)=50$ so $c=50$
$$v(t)=-4200t^2+2000t+50$$
We also know that $v(0.5)=0$
$$0=v(0.5)=-4200(0.5)^2+2000(0.5)+50=-1050+1000+50=0$$
Good.
Now $$x(t)=\int_0^tv(t)\,\mathrm dt=\int_0^t-4200t^2+2000t+50\,\mathrm dt=-1400t^3+1000t^2+50t+c$$
Again, we know $x(0)=0$ so $c=0$
$$x(t) = -1400t^3+1000t^2+50t$$
Last check: $$x(0.5)=100=-1400(0.5)^3+1000(0.5)^2+50(0.5)=-175+250+25=100$$
Check!
We're correct, now!
Since you just changed the problem, the new equation you get using this method is
$$a(t)=240t-160$$
If you don't decelerate at all, you will only cover $25$m in the half second, so you need to accelerate to cover the distance in that time. One approach would be to accelerate constantly for the half second, reach the $100$m point, and stop instantaneously. That gives $100=50 \cdot 0.5 + \frac 12 a (0.5)^2, a=8\cdot 75=600$m/s$^2$, or over $60 g$!?!?
