# Deriving the formula for distance

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.

UltrafastPED
Gold Member
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.

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 ?

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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.

jtbell
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.

1 person
Thanks guys !