### 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

\[v\left( t \right) = {v_0}t + at,\]

\[x\left( t \right) = {x_0} + {v_0}t + \frac{{a{t^2}}}{2}.\]

## 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 \(\large{\frac{\text{m}}{\text{s}}}\normalsize,\) 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 \(v = 2\sqrt{x}\,\left({\large{\frac{\text{m}}{\text{s}}}\normalsize}\right).\) 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 \(a\left( t \right) = \cos \large{\frac{{\pi t}}{6}}\normalsize\,\left( {\large{\frac{\text{m}}{{{\text{s}^2}}}}\normalsize} \right).\) Find the distance traveled by the particle for the \(3\text{rd}\) second.### 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 \(1\text{st}\) second it covers \({d_1}\) meters. What distance \({d_2}\) does the particle cover for the \(2\text{nd}\) second?### Example 7

A particle starts from rest and moves with an acceleration given by the equation \(a\left( t \right) = {e^t} – 1\,\left( {\large{\frac{\text{m}}{{{\text{s}^2}}}}\normalsize} \right).\) 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 \({v_0} = 6\,\left( {\large{\frac{\text{m}}{\text{s}}}\normalsize} \right)\) and acceleration \(5 – 2t\,\left( {\large{\frac{\text{m}}{{{\text{s}^2}}}}\normalsize} \right).\) Calculate the distance traveled by the particle for the first \(8\) seconds.### Example 9

A particle starts from rest and moves with constant acceleration \({a_1} = 5\,\left( {\large{\frac{\text{m}}{{{\text{s}^2}}}}\normalsize} \right)\) 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 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 \(\large{\frac{\text{m}}{\text{s}}}\normalsize,\) 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 \(v = 2\sqrt{x}\,\left({\large{\frac{\text{m}}{\text{s}}}\normalsize}\right).\) 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 \(a\left( t \right) = \cos \large{\frac{{\pi t}}{6}}\normalsize\,\left( {\large{\frac{\text{m}}{{{\text{s}^2}}}}\normalsize} \right).\) 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

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

### 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 \(a\,\left( {\large{\frac{\text{m}}{{{\text{s}^2}}}}\normalsize} \right)\). 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 \(a\left( t \right) = {e^t} – 1\,\left( {\large{\frac{\text{m}}{{{\text{s}^2}}}}\normalsize} \right).\) 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 \({v_0} = 6\,\left( {\large{\frac{\text{m}}{\text{s}}}\normalsize} \right)\) and acceleration \(5 – 2t\,\left( {\large{\frac{\text{m}}{{{\text{s}^2}}}}\normalsize} \right).\) 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 \({v_0} = 6\,\left( {\large{\frac{\text{m}}{\text{s}}}\normalsize} \right),\) 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.

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 \({a_1} = 5\,\left( {\large{\frac{\text{m}}{{{\text{s}^2}}}}\normalsize} \right)\) 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 \({v_0}\).

The velocity of the ball varies by the law

\[v\left( t \right) = {v_0} – gt,\]

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

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 \(\large{\frac{{{d_1}}}{{{d_2}}}}\normalsize = 2.\) 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,

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