Physics related antiderivatives problem I was wondering if I could get a little help with a calculus related word problem:
A car braked with a constant deceleration of 5 meters per second squared for 60 meters before stopping. How fast was the car traveling when the brakes were applied? I can't use the definite integral, only the antiderivative.
Thanks for any help!
 A: By definition the acceleration is $a = \dot v$, where $v$ is the velocity and $\dot{\phantom{v}}$ denotes derivative with respect to time. If $a$ is constant we can take the antiderivative to get
$$v(t) = at + v_0$$
where $v_0$ is the velocity at time $t=0$.
Now $v = \dot s$ where $s$ is the distance travelled. Using the above expression for $v$ we can again take the antiderivative and get
$$s(t) = \frac12at^2 + v_0t + s_0$$
where $s_0$ is the distance travelled at $t=0$.
Assume that the driver applies the brakes at time $t_0$ and that the car stops at time $t_1$. Then we have 
$$\Delta s = s(t_1) - s(t_0) = (\frac12at_1^2 + v_0t_1 + s_0) - (\frac12at_0^2 + v_0t_0 + s_0) = \frac12a(t_1^2-t_0^2) + v_0(t_1-t_0) \tag{1}$$
where $\Delta s$ is the distance travelled during braking, and also
$$0 = v(t_1) = at_1 + v_0 \tag{2}$$
In the exercise $\Delta s = 60\,\text{m}$ and $a = -5 \text{m/s}^2$ are given and we are asked for $v(t_0)$.
Using (2) we get $t_1 = -v_0/a$, which inserted into (1) gives
$$\Delta s = \frac12a((-v_0/a)^2-t_0^2) + v_0(-v_0/a-t_0) = -\frac12v_0^2/a - v_0t_0 - \frac12at_0^2 = -\frac12(v_0+at_0)^2/a = -\frac12v(t_0)^2/a$$
Thus,
$$v(t_0) = \pm\sqrt{2a\Delta s}$$
Now you can just insert the given values of $a$ and $\Delta s$.
I have in these calculations deliberately not taken $t_0=0$.
A: You just need the formula
$$V_1^2-V_2^2=2\gamma (X_1-X_2) $$
with $V_i =$ speed at position $X_i $ and $\gamma $ the constant acceleration.
thus
$$V_1^2-0=2.5.60$$
$$V_1=\sqrt {600} m/s$$
The formula comes from the integration of

$$\frac {d (V^2/2)}{dt}=V\frac {dV}{dt}=\gamma V=\gamma \frac {dX}{dt} $$

A: Let $travel_m = as^2+bs+c$ (s = time, travel_m = travelled distance)
$\frac{\text{d}^2y}{\text{d}s^2} as^2+bs+c$ (The second derivative of distance gives you the acceleration.)
$=2a=-5$
$a=-2.5$
$travel_m = -2.5s^2+bs+c$
At first, the travel distance was $0$, $c=0$.
$travel_m = -2.5s^2+bs$
$\frac{\text{d}y}{\text{d}s}-2.5s^2+bs$ (The first derivative of distance gives you the speed.)
$=-5s + b$, but at $s=0$, when the car braked, the speed was b, and they are asking for b.
When it stopped the speed was zero. To see when it stopped:
$-5s + b=0$ -> $\frac{b}{5}=s$
$-2.5s^2+bs = 60$
Substitute: $-\frac{5}{2}\times\frac{b^2}{25}+\frac{b^2}{5} = 60$ ->
$b^2=600$
$\boxed{b=10\sqrt6}$
