# Calculus

## Applications of the Derivative # Rectilinear Motion

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 $$x = 0.$$
1. Sketch the acceleration $$a$$ vs. time $$t$$ graph corresponding to this velocity vs. time graph;
2. Sketch the graph of position $$x$$ vs. time $$t$$ corresponding to the velocity vs. time graph;
3. 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.
1. Find the particle’s velocity;
2. Find the particle’s acceleration;
3. 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.
1. Find time $$t$$ when the acceleration is zero;
2. 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 $$x = 0.$$
1. Sketch the acceleration $$a$$ vs. time $$t$$ graph corresponding to this velocity vs. time graph;
2. Sketch the graph of position $$x$$ vs. time $$t$$ corresponding to the velocity vs. time graph;
3. 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:

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

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

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.
1. Find the particle’s velocity;
2. Find the particle’s acceleration;
3. 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.
1. Find time $$t$$ when the acceleration is zero;
2. 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 $$64\,\text{m}.$$