Proving kinematics formula using Calc?

**dandaman** · February 25th 2009, 07:22 PM

ok so say acceleration is constant.
$a=a$

the integral of a constant is
$\int a=at+C$

C in this case equals initial Velocity (V0)
so we have
$v=at+V0$

But how do i show that
$v^2=(v0)^2+2a(x-x0)$
using this type of integration?

Thanks for all the help in advance...

**Abu-Khalil** · February 25th 2009, 09:20 PM

Just replace $t$ from $x-x_0=v_0t+\frac{at^2}{2}\Rightarrow t=\frac{-v_0\pm\sqrt{v_0^2+2a(x-x_0)}}{a}.$

**arpitagarwal82** · February 25th 2009, 09:50 PM

Use this
a = dv/dt
=> a = (dv/dx) * (dx/dt)
=> a = v (dv/dt) since dx/dt = v

Now proceed.

**skeeter** · February 26th 2009, 05:04 AM

Originally Posted by dandaman

ok so say acceleration is constant.
$a=a$

the integral of a constant is
$\int a=at+C$

C in this case equals initial Velocity (V0)
so we have
$v=at+V0$

But how do i show that
$v^2=(v0)^2+2a(x-x0)$
using this type of integration?

Thanks for all the help in advance...

integrate velocity ...

$\Delta x = v_0t + \frac{1}{2}at^2$

using the velocity function ...

$ v_f = v_0 + at $

solve for t ...

$t = \frac{v_f - v_0}{a}$

substitute in the displacement equation to eliminate the parameter, t ...

$\Delta x = v_0 \left(\frac{v_f - v_0}{a}\right) + \frac{1}{2}a\left(\frac{v_f - v_0}{a}\right)^2$

$\Delta x = \frac{2v_0(v_f - v_0)}{2a} + \frac{(v_f - v_0)^2}{2a}$

$\Delta x = \frac{2v_0v_f - 2v_0^2 + v_f^2 - 2v_0v_f + v_0^2}{2a}$

$ \Delta x = \frac{v_f^2 - v_0^2}{2a} $

$ v_0^2 + 2a\Delta x = v_f^2 $

**dandaman** · February 26th 2009, 05:25 AM

ok now i see. Thanks to all.

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