## Introduction

Most of objects and phenomena in physics have motion. Therefore, to have a better understanding of how nature works, we should have an exact and clear definition of motion and ability of its computation.

Motion is defined as the change of an object's location with time which can be taken place in** linear, circular or projectile paths**. In a linear motion, whole parts of an object move in one direction or simply they have equal displacement and direction. Circular motion is the change of object's angular position. This is a movement along a circular path or circumference of a circle. Once an object moves or fired with an angle through space, we could call this type of motion as projectile one.

Previously, we defined **displacement** of an object and noticed it as the building block of the other kinematical concepts which describe the motion of a particle along an arbitrary path. Here, we study the concepts of velocity and acceleration in a problem-solution based approach with numerous solved examples. We believe that in this way the AP or college students can succeed more.

## Average velocity and speed (definition):

We need usually to know at what rate does the particle moves? To answer this question, we define an important quantity as follows: \[ Average\ speed=\frac{distance\ traveled}{elapsed\ time}=\frac{D}{\Delta t}\]

\[Average\ velocity=\frac{displacement}{elapsed\ time}\]

The average speed defines in terms of the distance traveled ($D$) so it is a scalar quantity. But the average velocity is a vector that points in the same direction as the displacement vector and defines as the change in position of a body divided by the time interval in which that changes occur. Average velocity is a more useful quantity than the other since it describes both how fast and in what direction the objects move. Therefore, in one direction, say $x$,the average velocity is defined as \[\vec{v}_{ave-x}=\frac{\Delta \vec{x}}{\Delta t}\]

The SI unit of velocity is meters per second ($\rm m/s$). Other common units include Kilometers per second ($\rm Km/h$), feet per second ($\rm ft/s$) and miles per hour ($\rm mi/h$).

Note that, in general, the magnitude of the average velocity and average speed are not equal in a given time interval. For example, if the above car moves from $A$ to $B$ in $4\,{\rm s}$ we have \[Average\ speed = \frac{10\,{\rm m}}{4\,{\rm s}}=2.5\ {\rm m/s}\] whereas \[Average\ velocity= \frac{\Delta \vec{x}}{\Delta t}=+\frac{4\,{\rm m}}{4\,{\rm s}}=+1\ {\rm m/s}\]

The positive indicates that the overall motion is in the positive $x$ direction.

**Example $1$:
A bird is flying $100\,{\rm m}$ due east at $10\,{\rm m/s}$ and then it turns around and flying west in $15\,{\rm s}$ at $20\,{\rm m/s}$. Find the average velocity and average speed during the overall time interval.**

**Solution**:

First we must find the total time. The first and second parts are done in $\Delta t_1=\frac{\Delta x}{v}=\frac{100}{10}=10\,{\rm s}$ and $\Delta t_2=15\, {\rm s}$. Thus the overall time of flying of the bird is $\Delta t_{tot}=10+15=25\,{\rm s}$. The distance traveled of the bird due west is $d_w=v\Delta t_2=20 \times 15=300\,{\rm m}$. Therefore, by definition of the average speed, we have

\[Average\ speed =\frac{(100+300)\,{\rm m}}{25\,{\rm s}}=16\ {\rm m/s}\]

As we can see from the figure, the displacement vector is $\Delta x=x_f-x_i=-200-0=-200\,{\rm m}$ so the average velocity is found as

\[\vec{v}_{ave}=\frac{\Delta \vec{x}}{\Delta t}=\frac{-200\,{\rm m}}{25\,{\rm s}}=-8\ {\rm m/s}\]

The negative indicates that the average velocity is towards the $–x$ direction.

To show the motion of a particle usually a position–versus-time ($x-t$) graph (or simply position graph) is used. In such plots, every point in the plane represents the coordinate and its corresponding time of a moving object. You should not be mistaken the path of motion with the curves in the $x-t$ graph. As you can see later, straight line movement of a car does not yield a straight line!. Further, time is specified on the horizontal axis. Using these graphs one can extract (or simplify) more information about the type of motion. Let's consider the velocity of a moving particle is constant during any successive equal-time intervals, then the $x-t$ graph is straight-line. This type of motion is called uniform motion. In this case, the average velocity is the slope of the position-versus-time graph. In one dimensional motion, this is simply

\[v_{ave-x}=\frac{\Delta x}{\Delta t}={\rm slope\ of\ the\ x-t\ graph}\]

Therefore, from the mathematical viewpoint, in the uniform motion, equal displacements occur during any successive equal-time interval.

If the velocity is not constant, the motion's graph is a curve as shown in the figure below (the actual path of the motion could be a straight line, this is a reason why the position graph is not a picture of the path!). In such a case, the object's average velocity during each particular time interval is the slope of the $x-t$ graph that connects the initial and final points in that interval $\Delta t$.

It is worth noting that in mathematics the slope of line connecting two points $A$ and $B$, in a plane, defines as $\tan$ of smaller angle with the $x$ axis. Therefore, we can relate the average speed or velocity (in which those two points get closer together) to the $\tan$ of smaller angle as

\[\text{average speed}=\text{slope of line $AB$}=\tan \alpha\]

## Instantaneous velocity

The average velocity is not a useful quantity for analyzing the non-uniform motions. Very often it is necessary to know the object’s velocity at a single instant of time. In this case, the instantaneous velocity can be defined. Instantaneous velocity is the limit of average velocity $\vec{v}_{ave}=\frac{\Delta \vec{x}}{\Delta t}$ as $\Delta t$ approaches zero. In

the language of calculus, this limit is called the derivative of $x$ with respect to $t$ and is written $\frac{dx}{dt}$. Thus in the mathematical form, we can state

\[\vec{v}_x=\lim_{\Delta t \to 0}\frac{\Delta \vec{x}}{\Delta t}=\frac{dx}{dt}\]

Therefore, the instantaneous velocity at time $t$ is the slope of the line that is tangent to the position-versus-time graph at the instance $t$. The time interval $\Delta t$ is always positive, so $v_x$ has the same sign as $\Delta x$.

In many textbooks, the velocity refers to the instantaneous velocity, so from now on, we adopt this convension.

**Example ($2$):
The position of a particle varies with time according to the equation $x(t) = 40-5t-5t^2$ where $x$ and $t$ are in $\rm m$ and $\rm s$, respectively. Find
($a$) The particle's average velocity between $t=1\,{\rm s}$ and $t=2\,{\rm s}$.
($b$) The velocity at $t=2\,{\rm s}$.**

**Solution:**

**(a)** By definition of the average velocity, we must first find the displacement during that interval.

\begin{eqnarray*}

x(t=2\,{\rm s})=40-5(2)-5(2)^2=10\ {\rm m}\\

x(t=1\,{\rm s})=40-5(1)-5(1)^2=30\ {\rm m}\\

\Delta x=x(t=2\,{\rm s})-x(t=1\,{\rm s})=10-30=-20\ {\rm m}

\end{eqnarray*}

Therefore, the average velocity becomes

\[v_{ave-x}=\frac{\Delta x}{\Delta t}=\frac{-20\,{\rm m}}{(2-1)\,{\rm s}}=-20\ {\rm m/s}\]

**(b)** The velocity is defined $v_x=\frac{dx}{dt}$

\[v_x=\frac{dx}{dt}=\frac{d}{dt} \left(40-5t-5t^2\right)=(-5-10\,t)\ {\rm m/s}\]

Evaluating the velocity at $t=2\,{\rm s}$ gives

\[v_x (t=2\,{\rm s})=-5-10(2)=-25\ {\rm m/s}\]

In both cases, the negatives indicate the particle moves in the $-x$ direction.

**Example $3$**:

**A student to arrive at her school, first walks $900\,{\rm m}$ to the north then $300\,{\rm m}$ toward the east. Finally walks $500\,{\rm m}$ along the south. The total time of walking is $5$ minutes.
($a$) Find the distance traveled by her?
($b$) Find the average velocity.**

**Solution**:

($a$) To find the distance traveled, one should add lengths of each part of the path with each other. So

\begin{eqnarray*}

\text{distance} &=& AB+BC+CD \\

&=& 900+300+500=1700\,{\rm m}

\end{eqnarray*}

($b$) As mentioned, average velocity is a vector quantity which depends on the displacement of an object between initial and final points. Thus, draw a vector which connects those points to each other as figure below (this is a simple geometrical problem)

\begin{eqnarray*}

\text{displacement}\,:\, \overrightarrow{AD} &=&\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}\\

|\overrightarrow{AD}|&=&\sqrt{(300)^2 +(400)^2 }=500\,{\rm m}\\

\end{eqnarray*}

In the first line, the definition of displacement as vector addition is represented and using Pythagoras theorem its magnitude is computed. Now, by definition, the magnitude of average velocity $|\vec{v}_{av}|$ read as

\begin{eqnarray*}

|\vec{v}_{av}| &=& \frac{\text{displacement}}{\text{total time}}\\

&=& \frac{500\,{\rm m}}{5 \times 60\,{\rm s}}\\

&=& \frac {5}{3}\,{\rm \frac m s}

\end{eqnarray*}

We can go further and find its direction. Recall that average velocity is a vector in the same direction as displacement vector. Therefore,

\begin{eqnarray*}

\theta &=& \tan^{-1} \left(\frac{\text{y-component of $\overrightarrow{AD}$}}{\text{x-component of $\overrightarrow{AD}$}} \right)\\

&=& \tan^{-1} \left(\frac{400}{300}\right)\\

&\cong& 53^\circ

\end{eqnarray*}

**Note**: in everyday language, we usually define velocity as the distance traveled divided by the elapsed time, say here $v=\frac{1700}{5\times 60}=\frac{17}{3}\,{\rm m/s}$. But in physics, there is distinction between these concepts.

**Practice $1$:
A car moves along the $x$ axis as following figure. Its position at instant $t_1=2\,{\rm s}$ is at $A$. It is positioned at $B$ at time $t_2=5\,{\rm s}$ then returns and passes through the point $C$ at $t_3=10\,{\rm s}$.
(a) What is the average speed and velocity of the car between the times $t_1$ and $t_2$?
b) Now find the above quantities in the time interval $t_1$ and $t_3$.**

You can find the solution at the problem 18 of the kinematics in one dimension section of the Exam Center.

## Acceleration

* Acceleration is defined as the rate of change of velocity of an object. That is, acceleration measures how quickly or slowly an object's velocity changes.* This change may be the result of the change in the magnitude or direction of the velocity or both of them so acceleration is a vector quantity. Thus if the magnitude of the velocity remains constant and only its direction varies, we can still have an accelerating motion (for example in the circular motions).

The

*average acceleration*during a particular time interval $\Delta t$ is defined as

\[\vec{a}_{ave}=\frac{\Delta \vec{v}}{\Delta t}\]

The $\text{SI}$ unit of acceleration is (meters per second) per second, abbreviated $\rm m/s^2$. The average acceleration like other kinematic parameters position $r$ and velocity $v$ is a vector and points in the same direction as velocity change $\Delta \vec{v}$.

If the velocity of a car changes from $0$ to $50\,{\rm km/h}$ in $10\,{\rm s}$, its average acceleration is found as

\[\vec{a}_{ave}=\frac{50\,(\rm Km/h)}{10\,{\rm s}}\]

This means that the velocity of the car increases by $5\, {\rm km/h}$ every $1\,{\rm s}$.

Instantaneous acceleration is the limit of the ratio $\frac{\Delta \vec{v}}{\Delta t}$ as $\Delta t$ approaches zero or in the mathematical language is the derivative of the velocity with respect to time.

\[\vec{a}=\lim_{\Delta t \to 0}\frac{\Delta \vec{v}}{\Delta t}=\frac{d\vec{v}}{dt}\]

The algebraic sign of the acceleration does not tell us about the object is speeding up or slowing down. To determine this, we must take into account the signs of both of acceleration and velocity. In one dimensional motion, say $x$ direction, if $\vec{v}_x$ and $\vec{a}_x$ are both positive, the speed of the object is increasing and it is speeding up. When $\vec{v}_x$ and $\vec{a}_x$ have opposite signs, the speed is decreasing.

**In summary, ****when $\vec{v}_x$ and $\vec{a}_x$ have the same sign, the speed is increasing and when they have opposite signs, the speed is decreasing which usually called decelerating motion**.

To find the type of motion, speeding up or slowing down, the exact way is that plot the $v-t$ graph and from it characterize the sign of acceleration and velocity as follows (see Examples below). In general, in constant(uniform) acceleration, the position and velocity graph are parabola and straight-line, respectively. But in case of non-uniform one, the $v-t$ graph is a smooth curve.

**Example ($4$):
A particle at time $t=3\, {\rm s}$ and at position $x=7\, {\rm m}$ has velocity $v=4\, {\rm m/s}$. At $t=7\, {\rm s}$ it is located at $x=-5\, {\rm m}$ with velocity $v=-2\, {\rm m/s}$. Find:
(a) The average velocity of the particle.
(b) The average acceleration of the particle.**

**Solution**

(a) By definition of the average velocity, we must first determine the displacement vector.so

\[\Delta \vec{x}=\vec{x}_f-\vec{x}_i=(-5\,{\rm m})-(7\, {\rm m})=-12\ {\rm m}\]

\[average\ velocity : \vec{v}_{ave}=\frac{\Delta \vec{x}}{\Delta t}=\frac{-12\, {\rm m}}{(7-3)\, {\rm s}}=\frac{-12}{4}=-3\ {\rm m/s}\]

The minus sign indicates that the direction of the motion of the particle points to the $–x$ axis.

(b) To find the average acceleration, first determine the change in the velocity of the particle during that interval time.

\[ \Delta \vec{v}=\vec{v}_f-\vec{v}_i=-2-4=-6\ {\rm m/s}\]

By definition we have

\[ \vec{a}_{ave}=\frac{\Delta \vec{v}}{\Delta t}=\frac{-6\, {\rm m/s}}{(7-3)\, {\rm s}}=\frac{-6}{4}=-1.5\ {\rm m/s^2}\]

**Example ($5$):
A bus accelerates uniformly from an initial velocity of $15\,{\rm m/s}$ to a final velocity of $7\,{\rm m/s}$ in $4$ seconds. Calculate the acceleration of the bus. Let the direction of motion of the bus be the positive direction**.

**Solution:**

Let's consider the positive direction of $+x$ axis as the direction of velocity of the car. We know that the subtracting the initial velocity from the final value divided by the time taken gets the average acceleration as

\begin{eqnarray*}

\vec{a}_{ave} &=&\frac{\Delta \vec{v}}{\Delta t}\\

&=& \frac{\vec{v}_f-\vec{v}_i}{t_f-t_i}=\frac{7-15}{4}\\

&=& -2\,{\rm m/s}

\end{eqnarray*}

As you can see, the direction of the car does not change (always toward the $+x$ direction) but the obtained acceleration is to the opposite side (the minus sign in front of it indicated), thus we here encounter a decelerating motion since the product of acceleration and velocity is negative i.e. $a_x v_x <0$.

**Practice $2$:
The position-versus-time graph of a moving object along a straight line is as follows. Discuss the type of motion, speeding up or slowing down, in the time interval $1\,{\rm s}$ to $3\,{\rm s}$?**

**Solution:**

To practice and see this solution go here.

**Example ($6$):
The equation of motion of a car in a straight line (in SI units) is as follows: $x(t)=t^3 - 6t^2 +9t$.
(a) What is the magnitude of the average acceleration in the interval $t_1=1\,{\rm s}$ and $t_2=2\,{\rm s}$.
(b) At what points, the car has changed its direction?
(c) Determine the type of motion along the entire path?**

**Solution:**

**(a)** The average acceleration is defined as the change of initial and final instantaneous velocity in a specific time interval $\Delta t$.

The variation of position with time is given as above. To obtain the desired quantity, we should take two steps, one calculate the velocity of the car at the given instants, next divide their difference by the time interval.

\begin{eqnarray*}

v(t)=\frac{dx}{dt}=\left(3\,t^{2}-12\,t+9\right)

\end{eqnarray*}

In above, we omitted the vector sign over the velocity $v$, since we are in straight line and the sign of velocity or acceleration imply the direction of the motion. Given the equation of velocity, $v(t)$, one can construct their initial and final values

\begin{eqnarray*}

v(t_1=1) &=& 3(1)^2-12(1)+9=0\,{\rm m/s}\\

v(t_2=2)&=&3(2)^2-12(2)+9=-3\,{\rm m/s}\\

\Delta v&=&v(t_2)-v(t_1)=-3-0=-3\,{\rm m/s}

\end{eqnarray*}

As noted above, the minus sign reveals the direction of motion toward the $-x$ axis. Therefore, the magnitude of average acceleration read

\begin{eqnarray*}

a_{ave}=\frac{\Delta v}{\Delta t}=\frac{-3}{2-1}=-3\,{\rm m/s^2 }

\end{eqnarray*}

Which as previously, the negative is direction of the acceleration. Because of $a_x v_x>0$, the motion is accelerating.

**(b)** A moving object to change direction, firstly should be stopped. To do this, simply find the roots of equation of velocity with time or roots of derivative of equation of motion.

\begin{eqnarray*}

v(t)=\frac{dx}{dt} &=&\frac{d}{dt}(t^3 )-\frac{d}{dt}(6t^2 )+\frac{d}{dt}(9t)=0\\

&=& 3t^2 -12t+9=0

\end{eqnarray*}

This is a quadratic equation, $a x^2 +b x+c=0$ with the following well-know solutions

\begin{eqnarray*}

x_{1,2}=\frac{-b\pm \sqrt{b^2 - 4ac}}{2a}

\end{eqnarray*}

Therefore,

\begin{eqnarray*}

t_{1,2} &=&\frac{-(-12)\pm\sqrt{(-12)^2 - 4(3)(9)}}{2(3)}\\

&=& t_{1,2}=\frac{12\pm\sqrt{144-108}}{6}\\

&\Longrightarrow& t_1=\frac{12+6}{6}=3\,{\rm s}\\

& & t_2=\frac{12-6}{6}=1\,{\rm s}

\end{eqnarray*}

For better understanding, we plotted $x-t$ graph of this motion. It is worth noting that since this plot is not a parabola so the motion has a non-uniform acceleration. Recall that the slope of $x-t$ graph is the velocity(speed). In a graph, extrema are where the first derivative of a function is zero. As one can see, in this plot, there are two points where the instantaneous velocity (first derivative of position with time) is zero. These points denotes when the car changes its direction or called **turning points**.

**(c)** As mentioned, first we should plot velocity graph. In part (a), velocity obtained by taking derivative of position equation $x(t)$. Here, we plotted the profile of $v(t)$ by a third-party drawing software like FX Draw or Google.

The car starts out moving to the right ($v>0$) with an initial velocity $v(t=0)=9\,{\rm m/s}$. The negative acceleration (the slope of $v-t$ graph indicates the acceleration) causes the speed to decrease (since the acceleration ($\vec{a}<0$) and velocity ($\vec{v}>0$) are in opposite directions) until the car reaches its first turning point (where the instantaneous velocity is zero, $v=0$) just before $t=1\,{\rm s}$. Now the car speeding up to the left ($v<0$ and $a<0$ so $a v>0$) until reaching maximum speed $v=3\,{\rm m/s}$ at $t=2\,{\rm s}$. In the time interval $t=2\,{\rm s}$ until $t=3\,{\rm s}$ when the car stops and changes its direction (the second turning point), the direction of motion is still to the left with an positive acceleration (slowing down). After the final turning point at $t=3\,{\rm s}$, the car moves to the right ($v> 0$) and the positive acceleration makes the velocity ever more positive (that is speeding up motion).

In summary, the motion in any time interval behaves as

\[

\left\{ \begin{array}{lll}

a_x<0 & v_x>0 & a_x v_x<0\, \text{(slowing down)} & 0\leq t\leq 1 \\

a_x<0 & v_x<0 & a_x v_x>0\, \text{(speeding up)} & 1\leq t\leq 2 \\

a_x>0 & v_x<0 & a_x v_x<0\, \text{(slowing down)} & 2\leq t\leq 3 \\

a_x>0 & v_x>0 & a_x v_x>0\, \text{(speeding up)} & t\geq 3 \\

\end{array} \right.

\]

26 July, 2017

Tags : Displacement, Examples of Acceleration and Deceleration, Average acceleration definition, Speed, Solved Examples of velocity-time graph , Instantaneous acceleration, kinematic