Deriving the formula for distance

[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.

 

[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.

 

[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 ?

 

[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.

 

[UltrafastPED]

You're doing too much work!

See http://physics.info/kinematics-calculus/

 

[jtbell]

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.

 

[DarthRoni]

Thanks guys !