# 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 (t) or x (t).

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

### 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.

$t = 20\,\text{s:}\;\;x = 20 \cdot 5 = 100\,\text{m};$
$t = 30\,\text{s:}\;\;x = 100 + 10 \cdot 5 \cdot \frac{1}{2}= 125\,\text{m};$
$t = 40\,\text{s:}\;\;x = 125 + 10 \cdot \left({-5}\right) \cdot \frac{1}{2}= 100\,\text{m};$
$t = 60\,\text{s:}\;\;x = 100 + 20 \cdot \left({-5}\right) = 0\,\text{m}.$

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

Consider two intervals:

$\left. 1 \right)\;0 \le t \le 30\,\text{sec},\;\;\left. 2 \right)\;30\,\text{sec} \lt t \le 60\,\text{sec}$

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.

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 $x\left( t \right) = - \frac{{{t^3}}}{6} + 2{t^2} - 1,$ 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 $v\left( t \right) = \cos t, u\left( t \right) = \sin 2t,$ 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 $$64\,\text{m}.$$