# 1. Applications of the Indefinite Integral

by M. Bourne

## Displacement from Velocity, and Velocity from Acceleration

High velocity train [Image source]

A very useful application of calculus is displacement, velocity and acceleration.

Recall (from Derivative as an Instantaneous Rate of Change) that we can find an expression for **velocity** by differentiating the expression for displacement:

`v=(ds)/(dt)`

Similarly, we can find the expression for the **acceleration** by differentiating the expression for velocity, and this is equivalent to finding the second derivative of the displacement:

`a=(dv)/dt=(d^2s)/(dt^2)`

It follows (since integration is the opposite process to differentiation) that to obtain the **displacement**,* *`s` of an object at time `t`* *(given the expression for velocity, `v`) we would use:

`s=int v\ dt`

Similarly, the **velocity** of an object at time `t` with acceleration `a`, is given by:

`v=inta\ dt`

### Example 1

A car starts from rest at `s=3\ "m"` from the origin and has acceleration at time `t` given by `a=2t-5\ "ms"^-2`. Find the velocity and displacement of the car at `t=4\ "s"`.

### Solution

We find the velocity using:

`v=int a\ dt`

So in this example we have:

`v=int(2t-5)\ dt`

`=t^2-5t+K`

When `t = 0`, `v = 0`, so `K =0`.

So the expression for velocity as a function of time is:

`v=t^2-5t\ text[m s]^-1`

When `t = 4`, `v = 4^2-5(4)=-4\ text[m s]^-1`

Now to find the displacement.

`s=int v\ dt`

Then

`s=int(t^2-5t)\ dt`

`=t^3/3 - (5t^2)/2 + K`

Now when `t = 0`, `s = 3`, so we substitute to obtain:

`3=0^3/3 - (5(0)t^2)/2 + K`

So `K=3` and therefore the general expression for `s` is:

`s=t^3/3 - (5t^2)/2 + 3`

When `t = 4`, `s = 4^3/3 - (5(4)^2)/2 + 3 = -15.67\ text[m]`

The graphs of the acceleration, velocity and displacement at time *t*, indicating the velocity and displacement at `t=4`.

### Example 2

A proton moves in an electric field such that its
acceleration (in cms^{-2}) is

`a = -20(1+2t)^-2`, where `t`is in seconds.

Find the velocity as
a function of time if *v* = 30 cms^{-1} when *t* = 0.

Answer

`v=int a\ dt`

So

`v=int(-20\ dt)/((1+2t)^2)`

Put `u = 1 + 2t` then `du = 2\ dt`, so `dt=(du)/2`

`v=int (-10\ du)/(u^2)`

`=int -10u^(-2)du`

`=10/u+K`

`=(10)/(1+2t)+K`

When `t = 0`, `v = 30`, so `K = 20`.

So the expression for velocity as a function of time is:

`v=((10)/(1+2t)+20)\ text[cm s]^-1`

Easy to understand math videos:

MathTutorDVD.com

Here are the graphs of the acceleration of the proton, and the velocity we found in Example 2.

The graphs of the acceleration and velocity at time *t*. Note `v(0)=30`.

### Example 3

A flare is ejected vertically upwards from the ground at 15 m/s. Find the height of the flare after 2.5 s.

Answer

The object has acting on it the force due to
gravity, so its acceleration is -9.8 ms^{-2}.

`v=int a dt`

`=int-9.8\ dt`

`=-9.8t+C`

Now at `t = 0`, the velocity = `15\ "ms"^-1`. So `C = 15`.

So the expression for velocity becomes:

`v=-9.8t+15`

Now, we need to find the displacement, so we integrate our expression for velocity:

`s=int v\ dt`

`=int(-9.8t + 15) dt`

`=-4.9t^2+15t+K`

Now, we know from the question that when `t = 0`, `s = 0`.

This gives `K = 0`.

So `s=-4.9t^2+15t`

At time `t = 2.5`, `s = 6.875\ "m"`.

## Displacement and Velocity Formulas

Using integration, we can obtain
the well-known expressions for displacement and velocity, given a
constant acceleration *a*,* *initial displacement zero,
and an initial velocity `v_0`:

`v=int a\ dt`

`v=at+K`

Since the velocity at `t=0` is `v_0`, we have `K=v_0`. So:

`v=v_0 + at`

Similarly, taking it another step gives:

`s=int v\ dt=int (v_0 + at)dt`

`s=v_0 t + (at^2)/2+C`

Since the displacement at `t=0` is `s=0`, we have `C=0`. So:

`s=v_0t+1/2at^2`

**Voltage across a Capacitor**

**Definition: **The current, *i *(amperes), in an electric circuit
equals the time rate of change of the **charge** *q*, (in
coulombs) that passes a given point in the circuit. We can write
this (with *t* in seconds) as:

`i=(dq)/(dt)`

By writing *i dt = dq* and **integrating**,
we have:

`q=inti\ dt`

The voltage, *V*_{C} (in
volts) across a capacitor with capacitance *C* (in farads)
is given by

`V_C=q/C`

It follows that

`V_C=1/Cinti\ dt`

You can see some more advanced applications of this at Applications of Ordinary Differential Equations.

### Example 4

The electric current (in mA) in a computer circuit as a function of time is `i = 0.3 − 0.2t`. What total charge passes a point in the circuit in `0.050` s?

Answer

The charge, *q*, is given by:

`q=inti\ dt`

`=int (0.3-0.2t) dt`

`=0.3t-0.1t^2+K`

At `t = 0`, `q = 0`, and this gives us `K = 0`.

So `q=0.3t-0.1t^2`

`[q]_[t=0.050]` `=0.3(0.050)-0.1(0.050)^2` `=0.015\ text[mC]`

(units are milli coulombs, as current `i`* *was in
`"mA"`.)

Easy to understand math videos:

MathTutorDVD.com

### Example 5

The voltage across an `8.50\ "nF"` capacitor in an FM receiver circuit
is initially zero. Find the voltage after `2.00\ μ"s"` if a current `i=0.042t`* *(in `"mA"`) charges the capacitor.

Answer

`V_C=1/C inti\ dt `

**Note:** `1\ "nF" = 10^-9"F"`; and `1\ μ"s" = 10^-6"s"`;

Also, `0.042t\ "mA" = 0.042 × 10^-3t\ "A"`

`V_C = (0.042 times 10^-3)/(8.5 times 10^-9)int t\ dt`

`=4.94 times 10^3 (t^2)/(2)+K`

`=2.47 times 10^3t^2 + K`

Now, we are told that when `t = 0`, `V_C= 0`.

So `K= 0`.

Thus

`V_C=2.47 times 10^3 t^2`

So when `t = 2.00\ μ"s"`, we have:

`V_C = 2.47 times 10^3(2 times 10^-6)^2`

`=9.882 times 10^-9`

`=9.88\ text[nV]`

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