# Deriving the formula for distance

1. Oct 14, 2013

### DarthRoni

I am currently redoing grade 11 physics and they have introduced the following formula,
$$\Delta d = v_1\Delta t + \frac{1}{2}a\Delta t^2$$
I am trying to find this equation using calculus.
So from what I understand, $\Delta d$ is the area under the curve for $v(\Delta t)$.
We can define $v(\Delta t) = v_1 + a\Delta t$ where $\Delta t = t_2 - t_1$
I can take the integral of $v(\Delta t)$ to find $\Delta d$ call it $D(\Delta t)$. Where, $D'(\Delta t) = v(\Delta t)$.
$$D(\Delta t) = \int v(\Delta t)\ d\Delta t \implies D(\Delta t) = \int (v_1 + a\Delta t)d\Delta t\implies D(\Delta t) = v_1\Delta t + \frac{1}{2}a\Delta t^2$$

and suppose I know $v(t)$ could I say:
$$\Delta d = \int_{t_1}^{t_2}v(t)dt = P(t_2) - P(t_1)\ where\ P'(t) = v(t)$$
$$\Delta d = D(\Delta t) \implies D(\Delta t) = P(t_2) - P(t_1)$$

Is this all correct ? Any feedback would be appreciated.

2. Oct 14, 2013

### UltrafastPED

We define velocity as v=dx/dt, and acceleration as a=dv/dt.

Start with the constant acceleration, a. One integration gives the change in velocity as v=at; the second gives the change in distance as d=1/2 at^2.

Then add in the expressions for an initial velocity, and an initial displacement - this gives the general result for constant acceleration.

3. Oct 14, 2013

### DarthRoni

So are you saying since $a = \frac{\Delta v}{\Delta t}\ and\ v = \frac{\Delta d}{\Delta t}$
$$\Delta v = \int (a)dt = at$$
$$\int (\Delta v) dt = \int (at)dt = \frac{1}{2}at^2$$
and since $v_1$ is constant
$$\int (v_1)dt = v_1t$$
and since $v(t) = v_1$ when $t = 0$, we have
$$\int v(t)dt = \int (v_1)dt + \int (\Delta v)dt$$
$$\implies \int v(t)dt = v_1t + \frac{1}{2}at^2$$

Is this what you mean? Where does initial displacement come into play? How would I proceed to use definite integrals ?
$$\int_{t_1}^{t_2}v(t)dt = (v_1t_2 + \frac{1}{2}at_2^2) -(v_1t_1 + \frac{1}{2}at_1^2)$$
does this work ?

Last edited: Oct 14, 2013
4. Oct 14, 2013

### DarthRoni

well I suppose all we did was define v(t) and took the integral. I'm still not sure what you mean by adding initial displacement.

5. Oct 14, 2013

### UltrafastPED

6. Oct 14, 2013

### Staff: Mentor

Here's another way to do it besides the one in the link UltrafastPED gave. Integrate the acceleration as an indefinite integral, so you add a constant of integration:
$$v=\int {a dt}\\ v = at + C$$

To find the constant of integration, you have to specify an "initial condition", namely that $v = v_0$ at t = 0:
$$v_0 = a \cdot 0 + C\\ C = v_0$$

Therefore $v = at + v_0$

You can probably do the second integration to get x, with the initial condition $x = x_0$ at t = 0.

7. Oct 14, 2013

### DarthRoni

Thanks guys !