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Velocity and Displacement

From, derivatives, we know that velocity, v, is the first derivative, and acceleration, a, is the second derivative. We can express this as


If we rearrange the equation to \(dv=a\,dt\), we can integrate in respect to t

\[v=\int a\,dt\]

If the acceleration is constant, we have \[v = at+c_1\]

A similar equation can be set up to find the displacement, s, in terms of time.

\[s=\int v\,dt\]

Example: During the initial stage of launching a spacecraft vertically, the acceleration \(a\,(in\,m/s^2)\) of the spacecraft is \(a=6t^2\). Find the height \(s\) of the spacecraft after \(6.0s\) if \(s=12m\) for \(t=0s\) and \(v=16\,m/3\) for \(t=2s\).


First, we find the equation of the veloctiy.

\begin{align} v&=\int 6t^2\,dt=2t^3+c_1 \\ 16&=2(2)^3+c_1 \end{align}

We solve and find that \(c_1 =0\), substitute \(c_1\) at \(t=2s\) back in and \(v=2t^3\).

Next we find the equation for the displacement.

\begin{align} s&=\int 2t^3\,dt=\frac{1}{2}t^4+c_2 \\ 12&=\frac{1}{2}(0)^4+c_2 \end{align}

We solve and find that \(c_2 =12\), substitute \(c_2\) at \(6.0s\) back in and \(s=\frac{1}{2}(6)^4+12=660m\).

Voltage Across a Capacitor

By definition, the current i in an electric circuit equals the time rate of change of the charge q (in coloumbs) that passes a given point in the circuit.



\[q=\int i\,dt\]

The voltage \(V_C\) across a capacitor with capacitance C is given by \(V_C=q/C\). Combining the equations the voltage is given by

\[V_C=\frac{1}{C} \int i\,dt\]

The measurements \(V_C\) is in volts, \(C\) is in farads, \(i\) in amperes, and \(t\) in seconds.

Example: A certain capacitor has \(100 V\) across it. At this instant, a current \(i=0.06t^{1/2}\) is sent through the circuit. After \(0.25s\), the voltage across the capacitor is \(140V\). What is the capacitance?

Substituting \(i=0.06t^{1/2}\) , we find that,

\begin{align} V_C &= \frac{1}{C}\int 0.06t^{\frac{1}{2}}dt \\ &= \frac{0.06}{C}t^{\frac{1}{2}}dt \\ &= \frac{0.04}{C}t^{\frac{3}{2}}+c_1 \end{align}

Now we solve for \(c_1\) with \(V_C=100\) at \(t=0\) and find that \(c_1=100V\). Thus,


We also know that \(V_C=140V\) at \(t=0.25s\)

\begin{align}140 &=\frac{0.04}{C}(0.25)^{\frac{3}{2}}+100 \\ 40&=\frac{0.04}{C}(0.125) \\ C&= 125\times 10^{-4} F=125\mu F\end{align}

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Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.
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