Rectilinear motion is a motion of a particle or object along a straight line.

Position is the location of object and is given as a function of time \(s\left( t \right)\) or \(x\left( t \right).\)

Velocity is the derivative of position:

\[{v = \frac{{dx}}{{dt}}.}\]

Acceleration is the derivative of velocity:

\[{a = \frac{{dv}}{{dt}}.}\]

The position and velocity are related by the Fundamental Theorem of Calculus:

\[{{\int\limits_{{t_1}}^{{t_2}} {v\left( t \right)dt} = \left. {x\left( t \right)} \right|_{{t_1}}^{{t_2}} }={ x\left( {{t_2}} \right) – x\left( {{t_1}} \right),}}\]

where \({t_1} \le t \le {t_2}.\) The quantity \(x\left( {{t_2}} \right) – x\left( {{t_1}} \right)\) is called a displacement. The displacement is represented by the area under the graph of the velocity \({v\left( t \right)}.\)

Similarly, since acceleration is the rate at which the velocity changes, we have

\[{{\int\limits_{{t_1}}^{{t_2}} {a\left( t \right)dt} = \left. {v\left( t \right)} \right|_{{t_1}}^{{t_2}} }={ v\left( {{t_2}} \right) – v\left( {{t_1}} \right),}}\]

where the quantity \(v\left( {{t_2}} \right) – v\left( {{t_1}} \right)\) is the net change in velocity in the time interval \({t_1} \le t \le {t_2}.\)

Speed \(\left| {v\left( t \right)} \right|\) is the absolute value of velocity, i.e. speed is always positive.

The average speed \({v_{av}}\) is defined as

\[{{{v_{av}} = \frac{{\text{total distance traveled}}}{{\text{total time}}}. }}\]

The total distance \(s\) a particle travels between time \({{t_1}}\) and time \({{t_2}}\) is given by

\[s = \int\limits_{{t_1}}^{{t_2}} {\left| {v\left( t \right)} \right|dt} .\]

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

The graph in Figure \(2\) below shows a particle’s velocity moving along a coordinate line. At \(t = 0,\) the position is \(x = 0.\)- Sketch the acceleration \(a\) vs. time \(t\) graph corresponding to this velocity vs. time graph;
- Sketch the graph of position \(x\) vs. time \(t\) corresponding to the velocity vs. time graph;
- Determine the average speed of the particle between \(t = 0\) and \(t = 60\,\text{sec}.\)

### Example 2

A particle moving on a line is at position \(x\left( t \right) = {t^3} – 9{t^2} + 24t – 5\) at time \(t\) where \(t\) is in seconds. At which time \(t\) (if any) does the particle change its direction?### Example 3

A particle moves along the \(x-\)axis according to the equation \(x\left( t \right) = 2{t^3} + 6{t^2} – 6t + 1,\) where time \(t \ge 0\) and is measured in seconds. Find the time when the particle’s velocity and acceleration are equal.### Example 4

The position function of a particle moving along the \(x-\)axis is given by \(x\left( t \right) = {t^3} – 4{t^2} + 5t – 2,\) \(t \ge 0.\) Find the open \(t-\)intervals when the particle is moving to the left.### Example 5

A particle moves along the \(x-\)axis so that its coordinate obeys the law \(x\left( t \right) = 2{t^2} + 4,\) where \(x\) is in meters and \(t\) is in seconds.- Find the particle’s velocity;
- Find the particle’s acceleration;
- Determine the average speed of the particle between \(t = 2\,\text{s}\) and \(t = 4\,\text{s}.\)

### Example 6

An object moves along the \(x-\)axis according to the law \(x\left( t \right) = – \large{\frac{{{t^3}}}{6}}\normalsize + 2{t^2} – 1,\) where \(x\) is in meters, \(t\) is in seconds.- Find time \(t\) when the acceleration is zero;
- Calculate the object’s velocity at this instant.

### Example 7

Find the integral expression that would result in the total distance traveled on the interval [0, 3] if the velocity is given by \(v\left( t \right) = {t^2} – 4.\)### Example 8

A particle is moving along the \(x\)-axis so that its position at time \(t \ge 0\) is given by the equation \(x\left( t \right) = t\ln t.\) Determine the acceleration of the particle when the velocity is zero.### Example 9

When two particles start at the origin with velocities \(v\left( t \right) = \cos t\) and \(u\left( t \right) = \sin 2t,\) how many times in the interval \(\left[ {0,2\pi } \right]\) will their speeds be equal?### Example 10

A particle moves along a straight line according to the equation \(x\left( t \right) = {t^3} – 6{t^2} + 5,\) where \(x\) is in meters, \(t\) is in seconds. Find the total distance traveled by the particle after 6 seconds.### Example 1.

The graph in Figure \(2\) below shows a particle’s velocity moving along a coordinate line. At \(t = 0,\) the position is \(x = 0.\)- Sketch the acceleration \(a\) vs. time \(t\) graph corresponding to this velocity vs. time graph;
- Sketch the graph of position \(x\) vs. time \(t\) corresponding to the velocity vs. time graph;
- Determine the average speed of the particle between \(t = 0\) and \(t = 60\,\text{sec}.\)

Solution.

\(1.\) Acceleration vs. time graph.

\[{a = \frac{{{v_2} – {v_1}}}{{{t_2} – {t_1}}} }={ \frac{{\left( { – 5} \right) – 5}}{{40 – 20}} }={ \frac{{ – 10}}{{20}} = – 0.5\frac{\text{m}}{{{\text{sec}^2}}}.}\]

\(2.\) Position vs. time graph.

\[{t = 20\,\text{s:}\;\;}\kern0pt{x = 20 \cdot 5 = 100\,\text{m};}\]

\[{t = 30\,\text{s:}\;\;}\kern0pt{x = 100 + 10 \cdot 5 \cdot \frac{1}{2}= 125\,\text{m};}\]

\[{t = 40\,\text{s:}\;\;}\kern0pt{x = 125 + 10 \cdot \left({-5}\right) \cdot \frac{1}{2}= 100\,\text{m};}\]

\[{t = 60\,\text{s:}\;\;}\kern0pt{x = 100 + 20 \cdot \left({-5}\right) = 0\,\text{m}.}\]

\(3.\) The average speed of the particle between \(t = 0\) and \(t = 60\,\text{sec}.\)

Consider two intervals:

\[{\left. 1 \right)\;0 \le t \le 30\,\text{sec},\;\;}\kern0pt{\left. 2 \right)\;30\,\text{sec} < t \le 60\,\text{sec}}\]

Though the particle returns to the initial position \(x = 0\,\text{m}\) at \(t = 60\,\text{sec},\) the total distance traveled from \(t = 0\) to \(t = 60\,\text{sec}\) is equal to

\[{s = \int\limits_0^{30} {\left| {v\left( t \right)} \right|dt} + \int\limits_{30}^{60} {{\left| {v\left( t \right)} \right|} dt} }={ 125 + 125 }={ 250\,\text{m}.}\]

The average speed is equal to

\[{{{v_{av}} = \frac{s}{{\Delta t}} = \frac{{250}}{{60}} }={ 4.17\,\frac{\text{m}}{\text{sec}}.}}\]

### Example 2.

A particle moving on a line is at position \(x\left( t \right) = {t^3} – 9{t^2} + 24t – 5\) at time \(t\) where \(t\) is in seconds. At which time \(t\) (if any) does the particle change its direction?Solution.

Find the particle’s velocity by differentiating position function:

\[{v\left( t \right) = x^\prime\left( t \right) }={ \left( {{t^3} – 9{t^2} + 24t – 5} \right)^\prime }={ 3{t^2} – 18t + 24 }={ 3\left( {{t^2} – 6t + 8} \right) }={ 3\left( {t – 2} \right)\left( {t – 4} \right).}\]

As you can see, the velocity becomes equal to zero at \(t = 2\,\text{s}\) and \(t = 4\,\text{s}.\) Hence, the particle changes its direction at the indicated times.

### Example 3.

A particle moves along the \(x-\)axis according to the equation \(x\left( t \right) = 2{t^3} + 6{t^2} – 6t + 1,\) where time \(t \ge 0\) and is measured in seconds. Find the time when the particle’s velocity and acceleration are equal.Solution.

We differentiate the position function successively to determine velocity and acceleration:

\[{v\left( t \right) = x^\prime\left( t \right) }={ \left( {2{t^3} + 6{t^2} – 6t + 1} \right)^\prime }={ 6{t^2} + 12t – 6,}\]

\[{a\left( t \right) = v^\prime\left( t \right) }={ \left( {6{t^2} + 12t – 6} \right)^\prime }={ 12t + 12.}\]

By equating \(v\) and \(a,\) we obtain

\[{6{t^2} + 12t – 6 }={ 12t + 12,}\]

\[6{t^2} = 18,\]

or

\[{t^2} = 3.\]

This equation has a positive root \(t = \sqrt 3. \)

Answer: \(t = \sqrt 3\,\text{s}.\)

### Example 4.

The position function of a particle moving along the \(x-\)axis is given by \(x\left( t \right) = {t^3} – 4{t^2} + 5t – 2,\) \(t \ge 0.\) Find the open \(t-\)intervals when the particle is moving to the left.Solution.

Find the particle’s velocity by differentiating the position function:

\[{v\left( t \right) = x^\prime\left( t \right) }={ \left( {{t^3} – 4{t^2} + 5t – 2} \right)^\prime }={ 3{t^2} – 8t + 5.}\]

Solve the quadratic equation:

\[{3{t^2} – 8t + 5 = 0,\;\;} \Rightarrow {D = {\left( { – 8} \right)^2} – 4 \cdot 3 \cdot 5 = 4,\;\;} \Rightarrow {{t_{1,2}} = \frac{{ – \left( { – 8} \right) \pm \sqrt 4 }}{6} }={ \frac{{8 \pm 2}}{6} }={ 1,\frac{5}{3}.}\]

Rewrite the velocity function in factored form:

\[{3{t^2} – 8t + 5 }={ 3\left( {t – 1} \right)\left( {t – \frac{5}{3}} \right).}\]

We see that the velocity is negative when \(1 \lt t \lt \frac{5}{3}\). In that time interval, the particle is moving to the left.

### Example 5.

A particle moves along the \(x-\)axis so that its coordinate obeys the law \(x\left( t \right) = 2{t^2} + 4,\) where \(x\) is in meters and \(t\) is in seconds.- Find the particle’s velocity;
- Find the particle’s acceleration;
- Determine the average speed of the particle between \(t = 2\,\text{s}\) and \(t = 4\,\text{s}.\)

Solution.

\(1.\) Particle’s velocity.

Take the derivative of \(x\left( t \right):\)

\[{v\left( t \right) = x^\prime\left( t \right) }={ \left( {2{t^2} + 4} \right)^\prime }={ 4t.}\]

Hence, the velocity of the particle is given by the equation

\[v = 4t\,\frac{\text{m}}{\text{s}}.\]

\(2.\) Particle’s acceleration.

To find acceleration, we differentiate the velocity function:

\[{a = v^\prime\left( t \right) }={ \left( {4t} \right)^\prime }={ 4\,\frac{\text{m}}{{{\text{s}^2}}}}.\]

As you can see, the particle moves with a constant acceleration.

\(3.\) Particle’s average speed.

First we determine the position of the particle at \(t = 2\,\text{s}\) and \(t = 4\,\text{s}.\)

\[{x\left( {{t_1}} \right) = x\left( 2 \right) }={ 2 \cdot {2^2} + 4 }={ 12\,\text{m}}\]

\[{x\left( {{t_2}} \right) = x\left( 4 \right) }={ 2 \cdot {4^2} + 4 }={ 36\,\text{m}}\]

Compute the average speed in the given time interval:

\[{{v_{av}} = \frac{{x\left( {{t_2}} \right) – x\left( {{t_1}} \right)}}{{{t_2} – {t_1}}} }={ \frac{{36 – 12}}{{4 – 2}} }={ 12\,\frac{\text{m}}{\text{s}}.}\]

### Example 6.

An object moves along the \(x-\)axis according to the law \(x\left( t \right) = – \large{\frac{{{t^3}}}{6}}\normalsize + 2{t^2} – 1,\) where \(x\) is in meters, \(t\) is in seconds.- Find time \(t\) when the acceleration is zero;
- Calculate the object’s velocity at this instant.

Solution.

The velocity is obtained by successive differentiation of \(x\) with respect to time \(t:\)

\[{v\left( t \right) = x^\prime\left( t \right) }={ \left( { – \frac{{{t^3}}}{6} + 2{t^2} – 1} \right)^\prime }={ – \frac{{{t^2}}}{2} + 4t.}\]

Similarly, to get the acceleration, we differentiate the velocity \(v\) with respect to time \(t:\)

\[{a\left( t \right) = v^\prime\left( t \right) }={ \left( { – \frac{{{t^2}}}{2} + 4t} \right)^\prime }={ – t + 4.}\]

Determine when the acceleration is equal to zero:

\[{a\left( t \right) = 0,\;\;} \Rightarrow {- t + 4 = 0,\;\;} \Rightarrow {t = 4\,\text{s}.}\]

Calculate the object’s velocity at \(t = 4:\)

\[{v(4) = – \frac{{{4^2}}}{2} + 4 \cdot 4 }={ 8\,\frac{\text{m}}{\text{s}}.}\]

### Example 7.

Find the integral expression that would result in the total distance traveled on the interval [0, 3] if the velocity is given by \(v\left( t \right) = {t^2} – 4.\)Solution.

To find the total distance we need to integrate the speed function, i.e. the absolute value of the velocity. Note that the velocity changes sign at \(t = 2.\) Therefore we split the interval \(\left[ {0,3} \right]\) into two intervals \(\left[ {0,2} \right]\) and \(\left[ {2,3} \right].\) The total distance \(s\) traveled by the particle on the interval \(\left[ {0,3} \right]\) is expressed in the form:

\[{s = \int\limits_0^2 {\left| {{t^2} – 4} \right|dt} }+{ \int\limits_2^3 {\left| {{t^2} – 4} \right|dt} .}\]

Given that the velocity is negative on the first interval and positive on the second interval, we get

\[{s = – \int\limits_0^2 {\left( {{t^2} – 4} \right)dt} }+{ \int\limits_2^3 {\left( {{t^2} – 4} \right)dt} .}\]

Rearranging the terms, we have

\[{s = \int\limits_2^3 {\left( {{t^2} – 4} \right)dt} }-{ \int\limits_0^2 {\left( {{t^2} – 4} \right)dt} .}\]

### Example 8.

A particle is moving along the \(x\)-axis so that its position at time \(t \ge 0\) is given by the equation \(x\left( t \right) = t\ln t.\) Determine the acceleration of the particle when the velocity is zero.Solution.

Find the particle’s velocity by differentiating the position function:

\[{v\left( t \right) = x^\prime\left( t \right) }={ \left( {t\ln t} \right)^\prime }={ 1 \cdot \ln t + t \cdot \frac{1}{t} }={ \ln t + 1.}\]

Continue differentiating to find the acceleration:

\[{a\left( t \right) = v^\prime\left( t \right) }={ \left( {\ln t + 1} \right)^\prime }={ \frac{1}{t}.}\]

The velocity is zero when time is

\[{v\left( t \right) = 0,\;\;} \Rightarrow {\ln t + 1 = 0,\;\;} \Rightarrow {\ln t = – 1,\;\;} \Rightarrow {t = \frac{1}{e}.}\]

Substituting this time value, we find the acceleration at this instant:

\[a = \frac{1}{{\frac{1}{e}}} = e.\]

### Example 9.

When two particles start at the origin with velocities \(v\left( t \right) = \cos t\) and \(u\left( t \right) = \sin 2t,\) how many times in the interval \(\left[ {0,2\pi } \right]\) will their speeds be equal?Solution.

We solve this problem graphically. Let’s draw the graphs of the speed functions \(\left| {v\left( t \right)} \right| = \left| {\cos t} \right|\) and \(\left| {u\left( t \right)} \right| = \left| {\sin 2t} \right|\) on the interval \(\left[ {0,2\pi } \right].\)

We see that the curves intersect \(4\) times on the interval \(\left[ {0,2\pi } \right].\)

### Example 10.

A particle moves along a straight line according to the equation \(x\left( t \right) = {t^3} – 6{t^2} + 5,\) where \(x\) is in meters, \(t\) is in seconds. Find the total distance traveled by the particle after 6 seconds.Solution.

To find the total distance traveled by a particle, we need to take the integral of of the speed \(\left| {v\left( t \right)} \right|:\)

\[s = \int\limits_{{t_1}}^{{t_2}} {\left| {v\left( t \right)} \right|dt} \]

Determine the particle’s velocity:

\[{v\left( t \right) = x^\prime\left( t \right) = \left( {{t^3} – 6{t^2} + 5} \right)^\prime }={ 3{t^2} – 12t }={ 3t\left( {t – 4} \right).}\]

We see that the velocity is negative for \(0 \lt t \lt 4\) and positive when \(t \gt 4.\) Hence, we split up the integral into the following two components:

\[{s = \int\limits_0^6 {\left| {v\left( t \right)} \right|dt} }={ \int\limits_0^4 {\left| {v\left( t \right)} \right|dt} }+{ \int\limits_4^6 {\left| {v\left( t \right)} \right|dt} .}\]

Given that the velocity is negative in the first integral and positive in the second, we obtain:

\[{s = – \int\limits_0^4 {v\left( t \right)dt} }+{ \int\limits_4^6 {v\left( t \right)dt}. }\]

This yields:

\[{s = – \left. {\left( {{t^3} – 6{t^2} + 5} \right)} \right|_0^4 }+{ \left. {\left( {{t^3} – 6{t^2} + 5} \right)} \right|_4^6 }={ – \left[ {\left( {64 – 96 + 5} \right) – 5} \right] }+{ \left[ {\left( {216 – 216 + 5} \right) – \left( {64 – 96 + 5} \right)} \right] }={ 32 + 32 }={ 64.}\]

So, the total distance traveled by the particle is equal to \(64\,\text{m}.\)