Calculus

Applications of Integrals

Applications of Integrals Logo

Distance, Velocity and Acceleration

Solved Problems

Click or tap a problem to see the solution.

Example 5

An object moves along a straight line with acceleration given by \[a\left( t \right) = 1 + \cos \left( {\pi t} \right).\] Assuming that the initial velocity at t = 0 is zero, find the total distance traveled over the first second.

Example 6

A particle starts moving from rest with a constant acceleration. For the 1st second it covers d1 meters. What distance d2 does the particle cover for the 2nd second?

Example 7

A particle starts from rest and moves with an acceleration given by the equation \[a\left( t \right) = {e^t} - 1\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right).\] Find the average speed of the particle over the interval from \({t_1} = 0\,\text{s}\) to \({t_2} = 2\,\text{s}.\)

Example 8

A particle starts with the initial velocity of \({v_0} = 6\,\left( {\frac{\text{m}}{\text{s}}} \right)\) and acceleration \(5 - 2t\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right).\) Calculate the distance traveled by the particle for the first \(8\) seconds.

Example 9

A particle starts from rest and moves with constant acceleration \({a_1} = 5\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right)\) for \({T_1} = 5\) seconds. It then decelerates uniformly for \({T_2} = 10\) seconds and stops. What is the distance covered by the particle?

Example 10

A ball thrown vertically travels double the distance in the \(1\text{st}\) second than in the \(2\text{nd}\) second. What is the maximum height achieved by the ball?

Example 5.

An object moves along a straight line with acceleration given by \[a\left( t \right) = 1 + \cos \left( {\pi t} \right).\] Assuming that the initial velocity at \(t = 0\) is zero, find the total distance traveled over the first second.

Solution.

First we compute the velocity of the object at time \(t.\) We use the formula

\[v\left( t \right) = {v_0} + \int\limits_0^t {a\left( u \right)du} ,\]

where \(u\) is a variable of integration. Hence,

\[v\left( t \right) = {v_0} + \int\limits_0^t {1 + \cos \left( {\pi u} \right)du} = {v_0} + \left. {\left[ {u + \frac{{\sin \left( {\pi u} \right)}}{\pi }} \right]} \right|_{u = 0}^{u = t} = 0 + \left( {t + \frac{{\sin \left( {\pi t} \right)}}{\pi }} \right) - \left( {0 + \frac{{\sin 0}}{\pi }} \right) = t + \frac{{\sin \left( {\pi t} \right)}}{\pi }.\]

Notice that the velocity \(v\left( t \right)\) is always positive for \(t \gt 0.\) Therefore, the total distance \(d\) traveled by the object for \(1\) sec is given by

\[d = \int\limits_0^1 {\left| {v\left( u \right)} \right|du} = \int\limits_0^1 {v\left( u \right)du} = \int\limits_0^1 {\left( {u + \frac{{\sin \left( {\pi u} \right)}}{\pi }} \right)du} = \left. {\left[ {\frac{{{u^2}}}{2} - \frac{{\cos \left( {\pi u} \right)}}{{{\pi ^2}}}} \right]} \right|_0^1 = \left( {\frac{1}{2} - \frac{{\cos \pi }}{{{\pi ^2}}}} \right) - \left( {0 - \frac{{\cos 0}}{{{\pi ^2}}}} \right) = \frac{1}{2} + \frac{2}{{{\pi ^2}}} \approx 0.70\,\left( \text{m} \right).\]

The distance \(d\) is measured in meters, if the acceleration \(a\) is measured in meters per second squared.

Example 6.

A particle starts moving from rest with a constant acceleration. For the \(1\text{st}\) second it covers \({d_1}\) meters. What distance \({d_2}\) does the particle cover for the \(2\text{nd}\) second?

Solution.

Let the acceleration of the particle be equal to \(a\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right)\). For motion from rest at constant acceleration, the velocity is given by the formula

\[v\left( t \right) = at.\]

The distance \({d_1}\) traveled by the particle for the \(1\text{st}\) second is written in the form

\[{d_1} = \int\limits_0^1 {v\left( t \right)dt} = \int\limits_0^1 {atdt} = \left. {\frac{{a{t^2}}}{2}} \right|_0^1 = \frac{a}{2}.\]

Similarly, we can write down the distance \({d_2}\) covered in the \(2\text{nd}\) second:

\[{d_2} = \int\limits_1^2 {v\left( t \right)dt} = \int\limits_1^2 {atdt} = \left. {\frac{{a{t^2}}}{2}} \right|_1^2 = 4a - \frac{a}{2} = \frac{{7a}}{2}.\]

Comparing both expressions, we find that

\[{d_2} = 7{d_1},\]

where \({d_1}\) and \({d_2}\) are measured in meters.

Example 7.

A particle starts from rest and moves with an acceleration given by the equation \[a\left( t \right) = {e^t} - 1\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right).\] Find the average speed of the particle over the interval from \({t_1} = 0\,\text{s}\) to \({t_2} = 2\,\text{s}.\)

Solution.

Integrating the acceleration function gives us the velocity function:

\[v\left( t \right) = \int {a\left( t \right)dt} = \int {\left( {{e^t} - 1} \right)dt} = {e^t} - t + C.\]

Since \({v_0} = 0,\) we have

\[v\left( 0 \right) = 0,\;\; \Rightarrow {e^0} - 0 + C = 0,\;\; \Rightarrow C = - 1.\]

Hence, the velocity function has the form:

\[v\left( t \right) = {e^t} - t - 1.\]

The average speed of an object is defined as the total distance traveled by the object divided by total time. So, we can write

\[\bar v = \frac{1}{{{t_2} - {t_1}}}\int\limits_{{t_1}}^{{t_2}} {\left| {v\left( t \right)} \right|dt} = \frac{1}{{{t_2} - {t_1}}}\int\limits_{{t_1}}^{{t_2}} {\left| {{e^t} - t - 1} \right|dt} = \frac{1}{{2 - 0}}\int\limits_0^2 {\left( {{e^t} - t - 1} \right)dt} = \frac{1}{2}\left. {\left( {{e^t} - \frac{{{t^2}}}{2} - t} \right)} \right|_0^2 = \frac{{{e^2} - 5}}{2} \approx 1.19\,\left( {\frac{\text{m}}{\text{s}}} \right)\]

Example 8.

A particle starts with the initial velocity of \({v_0} = 6\,\left( {\frac{\text{m}}{\text{s}}} \right)\) and acceleration \(5 - 2t\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right).\) Calculate the distance traveled by the particle for the first \(8\) seconds.

Solution.

We calculate the velocity function from the acceleration function:

\[v\left( t \right) = \int {a\left( t \right)dt} = \int {\left( {5 - 2t} \right)dt} = 5t - {t^2} + C.\]

To find \(C,\) we use the initial condition \({v_0} = 6\,\left( {\frac{\text{m}}{\text{s}}} \right),\) so we get

\[v\left( t \right) = 5t - {t^2} + 6.\]

Notice that the velocity is equal to zero when \(t =6\) seconds. Hence, when calculating the distance, we split the interval of integration into two intervals where the velocity has a constant sign.

Graphs of acceleration, velocity, position and total distance traveled by a particle.
Figure 2.

Thus, the total distance traveled by the particle is given by

\[d = \int\limits_0^8 {\left| {v\left( t \right)} \right|dt} = \int\limits_0^8 {\left| {5t - {t^2} + 6} \right|dt} = \int\limits_0^6 {\left( {5t - {t^2} + 6} \right)dt} - \int\limits_6^8 {\left( {5t - {t^2} + 6} \right)dt} = \left. {\left( {\frac{{5{t^2}}}{2} - \frac{{{t^3}}}{3} + 6t} \right)} \right|_0^6 - \left. {\left( {\frac{{5{t^2}}}{2} - \frac{{{t^3}}}{3} + 6t} \right)} \right|_6^8 = 2 \cdot 54 - \frac{{112}}{3} = \frac{{212}}{3} \approx 70.67\,\text{m}\]

Example 9.

A particle starts from rest and moves with constant acceleration \({a_1} = 5\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right)\) for \({T_1} = 5\) seconds. It then decelerates uniformly for \({T_2} = 10\) seconds and stops. What is the distance covered by the particle?

Solution.

Determine the velocity \({v_1}\) at the moment \(t = {T_1}:\)

\[{v_1} = {a_1}t = {a_1}{T_1} = 5 \cdot 5 = 25\,\left( {\frac{\text{m}}{\text{s}}} \right).\]

On the second stage, the particle's velocity varies according to the equation

\[v = {v_1} + {a_2}t,\]

where \(0 \le t \le {T_2}.\) Since \(v\left( {{T_2}} \right) = 0,\) we can easily find the value of acceleration \({a_2}:\)

\[{a_2} = - \frac{{{v_1}}}{{{T_2}}} = - \frac{{{a_1}{T_1}}}{{{T_2}}} = - \frac{{25}}{{10}} = - 2.5\,\left( {\frac{m}{{{s^2}}}} \right)\]

The value of \({a_2}\) is negative as the particle is decelerating.

Let's now move on to the calculation of distances. On the first stage, the distance traveled by the particle is equal to

\[{d_1} = \int\limits_0^{{T_1}} {{a_1}tdt} = {a_1}\int\limits_0^{{T_1}} {tdt} = \left. {\frac{{{a_1}{t^2}}}{2}} \right|_0^5 = \frac{{5 \cdot {5^2}}}{2} = 62.5\,\text{m}\]

The distance traveled on the second stage is given by

\[{d_2} = \int\limits_0^{{T_2}} {\left( {{v_1} + {a_2}t} \right)dt} = \int\limits_0^{{T_2}} {\left( {{v_1} + {a_2}t} \right)dt} = \left. {\left( {{v_1}t + \frac{{{a_2}{t^2}}}{2}} \right)} \right|_0^{10} = 25 \cdot 10 - \frac{{2,5 \cdot {{10}^2}}}{2} = 125\,\text{m}\]

Hence, the total distance covered by the particle is

\[d = {d_1} + {d_2} = 62.5 + 125 = 187.5\,\text{m}\]

Example 10.

A ball thrown vertically travels double the distance in the \(1\text{st}\) second than in the \(2\text{nd}\) second. What is the maximum height achieved by the ball?

Solution.

Suppose that the ball is thrown upward with an initial velocity of \({v_0}.\)

A ball thrown vertically with an initial velocity of v0 achieves the height H.
Figure 3.

The velocity of the ball varies by the law

\[v\left( t \right) = {v_0} - gt,\]

where \(g\) is the acceleration due to gravity: \(g \approx 9.8\,\left( {\frac{\text{m}}{{{\text{s}^2}}}} \right)\)

The distance traveled in the \(1\text{st}\) second is

\[{d_1} = \int\limits_0^1 {\left( {{v_0} - gt} \right)dt} = \left. {\left( {{v_0}t - \frac{{g{t^2}}}{2}} \right)} \right|_0^1 = {v_0} - \frac{g}{2}.\]

Respectively, the distance covered for the \(2\text{nd}\) second is given by

\[{d_2} = \int\limits_1^2 {\left( {{v_0} - gt} \right)dt} = \left. {\left( {{v_0}t - \frac{{g{t^2}}}{2}} \right)} \right|_1^2 = \left( {2{v_0} - 2g} \right) - \left( {{v_0} - \frac{g}{2}} \right) = {v_0} - \frac{{3g}}{2}.\]

We can find \({v_0}\) from the condition \(\frac{{{d_1}}}{{{d_2}}} = 2.\) So

\[\frac{{{v_0} - \frac{g}{2}}}{{{v_0} - \frac{{3g}}{2}}} = 2,\;\; \Rightarrow {v_0} - \frac{g}{2} = 2\left( {{v_0} - \frac{{3g}}{2}} \right),\;\; \Rightarrow {v_0} - \frac{g}{2} = 2{v_0} - 3g,\;\; \Rightarrow {v_0} = \frac{{5g}}{2}.\]

The maximum height is determined by the equation

\[H = \frac{{v_0^2}}{{2g}}.\]

Hence,

\[H = \frac{{{{\left( {\frac{{5g}}{2}} \right)}^2}}}{{2g}} = \frac{{25{g^\cancel{2}}}}{{8\cancel{g}}} = \frac{{25g}}{8} \approx 30.7\,\left( \text{m} \right)\]
Page 1 Page 2