Applications of Integrals

Applications of Integrals Logo

Distance, Velocity and Acceleration

Basic Relationships

Recall from our study of derivatives that if x (t) 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

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 \(5\,\text{sec}.\)


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.


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 \(t = 5\,\text{s}.\)


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

\[\Delta x = \int\limits_{{t_1}}^{{t_2}} {v\left( t \right)dt} = \int\limits_0^5 {\left( {6 - 2t} \right)dt} = \left. {\left( {6t - {t^2}} \right)} \right|_0^5 = 5\,\text{m}.\]
Velocity, displacement and distance traveled by an object when the velocity changes its sign.
Figure 1.

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.


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.

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