# Calculus

## Applications of Integrals # Distance, Velocity and Acceleration

## Basic Relationships

Recall from our study of derivatives that if $$x\left( t \right)$$ is the position of an object moving along a straight line at time $$t,$$ then the velocity of the object is.

$v\left( t \right) = \frac{{dx}}{{dt}},$

and the acceleration is given by

$a\left( t \right) = \frac{{dv}}{{dt}} = \frac{{{d^2}x}}{{d{t^2}}}.$

Using the integral calculus, we can calculate the velocity function from the acceleration function, and the position function from the velocity function.

The Fundamental Theorem of Calculus says that

$\int\limits_{{t_1}}^{{t_2}} {a\left( t \right)dt} = \left. {v\left( t \right)} \right|_{{t_1}}^{{t_2}} = v\left( {{t_2}} \right) - v\left( {{t_1}} \right).$

Similarly, the difference between the position at time $${{t_1}}$$ and the position at time $${{t_2}}$$ is determined by the equation

$\int\limits_{{t_1}}^{{t_2}} {v\left( t \right)dt} = \left. {x\left( t \right)} \right|_{{t_1}}^{{t_2}} = x\left( {{t_2}} \right) - x\left( {{t_1}} \right).$

If the object moves from the position $$x\left( {{t_1}} \right)$$ to the position $$x\left( {{t_2}} \right),$$ the change $$x\left( {{t_2}} \right) - x\left( {{t_1}} \right)$$ is called the displacement of the object:

$\Delta x = x\left( {{t_2}} \right) - x\left( {{t_1}} \right).$

To find the total distance traveled by the object between time $${{t_1}}$$ and time $${{t_2}},$$ we need to compute the integral of $$\left| {v\left( t \right)} \right|:$$

$d = \int\limits_{{t_1}}^{{t_2}} {\left| {v\left( t \right)} \right|dt} .$

## Constant Acceleration

Suppose that an object is moving along a straight line with the constant acceleration $$a.$$ At time $${t_1} = 0,$$ the object has an initial velocity $${v_0}$$ and an initial position $${x_0}.$$

The velocity and position of the object at time $$t$$ are given by the equations

$v\left( t \right) = {v_0} + at,$
$x\left( t \right) = {x_0} + {v_0}t + \frac{{a{t^2}}}{2}.$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

The velocity of an object is given by the equation $v\left( t \right) = \sqrt {4 + t} ,$ where the velocity $$v$$ is measured in $$\frac{\text{m}}{\text{s}},$$ the time $$t$$ is measured in seconds. Find the distance traveled by the object for the first $$5\,\text{sec}.$$

### Example 2

A particle moves along a straight line in the positive direction of the $$x-$$axis with velocity given by the equation $v = 2\sqrt{x}\,\left({\frac{\text{m}}{\text{s}}}\right).$ Assuming that $$x\left( {t = 0} \right) = 0,$$ find the time $$t$$ the particle takes to cover the first $$100\,\text{m}$$ of the path.

### Example 3

Starting at time $$t = 0,$$ an object moves along a straight line with the velocity $v\left( t \right) = 6 - 2t.$ Calculate the displacement and distance traveled by the object at time $$t = 5\,\text{s}.$$

### Example 4

A particle starts from rest with an acceleration $$a\left( t \right)$$ which varies according to the equation $a\left( t \right) = \cos \frac{{\pi t}}{6}\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right).$ Find the distance traveled by the particle for the $$3\text{rd}$$ second.

### Example 1.

The velocity of an object is given by the equation $v\left( t \right) = \sqrt {4 + t} ,$ where the velocity $$v$$ is measured in $$\frac{\text{m}}{\text{s}},$$ the time $$t$$ is measured in seconds. Find the distance traveled by the object for the first $$5\,\text{sec}.$$

Solution.

Given that the velocity is positive for $$t \gt 0,$$ the total distance traveled for the time interval $$\left[ {0,t} \right)$$ is expressed by the integral:

$d\left( t \right) = \int\limits_0^t {\left| {v\left( u \right)} \right|du} = \int\limits_0^t {v\left( u \right)du} = \int\limits_0^t {\sqrt {4 +u} du} ,$

where $$u$$ is the inner variable which has no impact on the computation of the integral.

Integration yields:

$d\left( t \right) = \left. {\frac{{2\sqrt {{{\left( {4 + u} \right)}^3}} }}{3}} \right|_0^t = \frac{2}{3}\left[ {\sqrt {{{\left( {4 + t} \right)}^3}} - 8} \right].$

Substituting $$t = 5,$$ we have

$d\left( {t = 5\,\text{s}} \right) = \frac{2}{3}\left[ {\sqrt {{{\left( {4 + 5} \right)}^3}} - 8} \right] = \frac{2}{3} \cdot 19 = \frac{{38}}{3} \approx 12.7\,\text{m}$

### Example 2.

A particle moves along a straight line in the positive direction of the $$x-$$axis with velocity given by the equation $v = 2\sqrt{x}\,\left({\frac{\text{m}}{\text{s}}}\right).$ Assuming that $$x\left( {t = 0} \right) = 0,$$ find the time $$t$$ the particle takes to cover the first $$100\,\text{m}$$ of the path.

Solution.

The equation of motion of the particle has the form

$v = \frac{{dx}}{{dt}} = 2\sqrt x .$

We have a simple differential equation that describes the particle's position as a function of time. Separating the variables and integrating both sides yields:

$\frac{{dx}}{{\sqrt x }} = 2dt,\;\; \Rightarrow \int {\frac{{dx}}{{\sqrt x }}} = 2\int {dt} ,\;\; \Rightarrow 2\sqrt x = 2t + C,\;\; \Rightarrow \sqrt x = t + C.$

It follows from the initial condition $$x\left( {t = 0} \right) = 0$$ that the $$C = 0.$$ Hence, the particle moves according to the law:

$x = {t^2}.$

It is easy to see that $$t = 10\,\text{s}$$ when $$x = 100\,\text{m}.$$

### Example 3.

Starting at time $$t = 0,$$ an object moves along a straight line with the velocity $v\left( t \right) = 6 - 2t.$ Calculate the displacement and distance traveled by the object at time $$t = 5\,\text{s}.$$

Solution.

Let's first compute the displacement $$\Delta x$$ of the object when $$t = 5\,\text{s}.$$ Integrating the velocity expression, we obtain

$\Delta x = \int\limits_{{t_1}}^{{t_2}} {v\left( t \right)dt} = \int\limits_0^5 {\left( {6 - 2t} \right)dt} = \left. {\left( {6t - {t^2}} \right)} \right|_0^5 = 5\,\text{m}.$

Notice that the velocity changes sign at $${t_1} = 3.$$ Therefore, to calculate the distance $$d$$ traveled by the object we split the initial interval $$\left[ {0,5} \right]$$ into two subintervals $$\left[ {0,3} \right]$$ and $$\left[ {3,5} \right].$$ This yields

$d = \int\limits_{{t_1}}^{{t_2}} {\left| {v\left( t \right)} \right|dt} = \int\limits_0^5 {\left| {6 - 2t} \right|dt} = \int\limits_0^3 {\left| {6 - 2t} \right|dt} + \int\limits_3^5 {\left| {6 - 2t} \right|dt} = \int\limits_0^3 {\left( {6 - 2t} \right)dt} - \int\limits_3^5 {\left( {6 - 2t} \right)dt} = \left. {\left( {6t - {t^2}} \right)} \right|_0^3 - \left. {\left( {6t - {t^2}} \right)} \right|_3^5 = \left[ {\left( {18 - 9} \right) - 0} \right] - \left[ {\left( {30 - 25} \right) - \left( {18 - 9} \right)} \right] = 9 + 4 = 13\,\text{m}.$

### Example 4.

A particle starts from rest with an acceleration $$a\left( t \right)$$ which varies according to the equation $a\left( t \right) = \cos \frac{{\pi t}}{6}\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right).$ Find the distance traveled by the particle for the $$3\text{rd}$$ second.

Solution.

Given that the initial velocity is zero: $${v_0} = 0,$$ we determine the velocity equation:

$v\left( t \right) = \int {a\left( t \right)dt} = \int {\cos \frac{{\pi t}}{6}dt} = \frac{6}{\pi }\sin \frac{{\pi t}}{6}.$

The distance traveled in the $$3\text{rd}$$ second is

$d = \int\limits_2^3 {v\left( t \right)dt} = \int\limits_2^3 {\frac{6}{\pi }\sin \frac{{\pi t}}{6}dt} = \frac{{{6^2}}}{{{\pi ^2}}}\left. {\left( { - \cos \frac{{\pi t}}{6}} \right)} \right|_2^3 = \frac{{36}}{{{\pi ^2}}}\left( { - \cos \frac{\pi }{2} + \cos \frac{\pi }{3}} \right) = \frac{{36}}{{{\pi ^2}}}\left( {0 + \frac{1}{2}} \right) = \frac{{18}}{{{\pi ^2}}} \approx 1.82\,\text{m}$

See more problems on Page 2.