Speed (v): rate of change of distance
Speed is a
Velocity ( $\vec{v})$ : rate of change in position
Velocity is a
Ex: A student travels $11 \mathrm{~m}$ north and then turns around and travels $25 \mathrm{~m}$ south. If the total time of travel is $12 \mathrm{~s}$, find:
a) The student’s average speed.
▶️Answer/Explanation
$\begin{array}{r}v=\frac{d}{t}=\frac{(11+25)}{12}=\frac{36 \mathrm{~m}}{12 \mathrm{~s}} \\ =3.0 \mathrm{~m} / \mathrm{s}\end{array}$
Question
b) The student’s average velocity.
▶️Answer/Explanation
$\vec{d}= – 14 \mathrm{m}$
$\begin{aligned} \vec{V}=\frac{d}{t}=-\frac{14}{12} & =-1.16667& =-1.7 m/s\end{aligned}$
Ex.
1) How long does it take a car traveling at $45 \mathrm{~km} / \mathrm{h}$ to travel $100.0 \mathrm{~m}$ ?
▶️Answer/Explanation
$\begin{array}{r}45 \mathrm{~km} / \mathrm{h} \div 3.6 \\ =12.5 \mathrm{m/s}\end{array}$
$t. V=\frac{d}{t} \cdot t$
$\frac{v \cdot t}{v}=\frac{d}{v}$
$t=\frac{d}{v}$
=-8.0s
Ex
2) How far does a skateboarder travel in $22 \mathrm{~s}$ if his average velocity is $12.0 \mathrm{~m} / \mathrm{s}$ ?
▶️Answer/Explanation
$V.t=\frac{d}{t} \cdot t$
$d=v \cdot t=(12.0)(22)=264$
3 – Uniform Accelerated Motion
(1) Acceleration: rate of change of velocity
- Acceleration is a vecter_______
Anytime an object’s…velocity changes, it is accelerating
$\vec{a}=\frac{\Delta \vec{v}}{t}$
Where:
$\vec{a}$ =accelevation
$\Delta$ = detla= change in
$=$ final-initial.
$\Delta \vec{v}$ = Velocity
t= time
1) A sprinter starts from rest and reaches a speed of $12 \mathrm{~m} / \mathrm{s}$ in $4.25 \mathrm{~s}$. Find his acceleration.
▶️Answer/Explanation
$V_i=0$ $V_f=12 \mathrm{~m} / \mathrm{s}$
$a=\frac{\Delta v}{t}=\frac{v_f-v_i}{t}$
$\begin{aligned} & =\frac{12}{4.25}=2.824 \\ & =2.8\end{aligned}$
2) A car starts from rest and accelerates at $15 \mathrm{~m} / \mathrm{s}^2$ for $3.0 \mathrm{~s}$. What is its top speed?
▶️Answer/Explanation
$\vec {a}= 15 \mathrm{~m} / \mathrm{s}^2$
$a=\frac{\Delta v}{t} \quad a=\frac{v_f-V_i}{t}$
$a.t =\frac{V_f}{t}.t$
$V_f=a \cdot t=(15)(3.0)=45 \mathrm{~m} / \mathrm{s}$
$V_i=8.0 \mathrm{~m} / \mathrm{s}$
$V_f=36.0 \mathrm{~m} / \mathrm{s}$
$\vec {a}=3.5 \mathrm{~m} / \mathrm{s}^2$
$a.t=\frac{\Delta v}{t}.t$
$\frac{a t}{a}=\frac{\Delta V}{a}$
$\begin{aligned} t & =\frac{\Delta v}{a} \\ & =\frac{v_f-v_i}{a}\end{aligned}$
$=\frac{(36.0-8.0)}{3.5}m/s$
$=8.0 \mathrm{~s}$
Question
Remember that all vectors include...directron.
- Up or to the right are:+
- Down or to the left are:–
An object’s velocity and acceleration can…
be in difterent directions
5 – The Big 3 Kinematics
If an object is accelerating then the formula:
$\vec{v}=\frac{\vec{d}}{t}$ Gives us only the average velocity
We can also find average velocity using:
$V_{\text {average }}=\frac{V+V_0}{2}$
In order to solve problems with uniform acceleration we need to use 3 formulae. These 3 formulae use the variables:
$\begin{aligned} & \left.\begin{array}{l}v=\text { final velocity } \\ v_0=\text { initial velocity }\end{array}\right\} \text { vectors } \\ & \left.\begin{array}{l}a=\text { acceleration } \\ d=\text { displacecent }\end{array}\right\}\text { vectors } \\ & t=\text { time } \leftarrow \text { scalar } \\ & \end{aligned}$
1) $\quad V=V_0+a t$
Ex: A squad car traveling at $7.0 \mathrm{~m} / \mathrm{s}$ East speeds up to $22.0 \mathrm{~m} / \mathrm{s}$ East in $1.7 \mathrm{~s}$. What is its acceleration?
$\overrightarrow{7.0 \mathrm{~m} / \mathrm{s}} \quad \overrightarrow{22.0 \mathrm{~m} / \mathrm{s}}$
$\begin{array}{lc}V=22.0 \mathrm{~m} / \mathrm{s} & V=V_0+a t \\ V_0=7.0 \mathrm{~m} / \mathrm{s} & -V_0-V_0 \\ a= & \frac{V-V_0}{t}=\frac{a t}{t} \\ d= & a=\frac{V-V_0}{t}=\frac{(22.0-7.0)}{1.7} \\ t=1.7 \mathrm{~s} & =0.021=8.8 \mathrm{m/s}^2\end{array}$
$d=v_0 t+\frac{1}{2} a t^2$
Example:
A sprinter starts from rest and accelerates uniformly. He travels $40.0 \mathrm{~m}$ in $4.82 \mathrm{~s}$, what was his average acceleration?
$V=$
$V_0=0$ $d=v_0 t+\frac{1}{2} a t^2$
$a=$ ? $2 d=\left(\frac{1}{2} a t^2\right) \quad 2 d=a t^2$
d=40.0m $2 d=\left(\frac{1}{2} a t^2\right) \quad \frac{2 d}{t^2}=\frac{a t^2}{t^2}$
t=-4.82s $a=\frac{2 d}{t^2}=\frac{2(40.0)^{\top}}{4.82^2}$
$=3.443=3.44 \mathrm{~m} / \mathrm{s}^2$
$V^2=V_0^2+2 a d$
Example:
A banana boat accelerates from $15.0 \mathrm{~km} / \mathrm{h}$ at $2.00 \mathrm{~m},{ }^2$. How far has it traveled when it reaches $30.0 \mathrm{~km} / \mathrm{h}$ ?
▶️Answer/Explanation
$\begin{aligned} & V=30.0 \mathrm{ka} / \mathrm{h} \div 3.6=8.333 \mathrm{m/s} \\ & V_0=15.0 \mathrm{k} / \mathrm{h} \div 3.6=4.167 \mathrm{m/s} \\ & a=2.00 \mathrm{n} / \mathrm{s}^2 \\ & d=? \\ & t=\end{aligned}$
$\frac{v^2-v_0^2}{2 a}=\frac{2 a d}{2 a}$
$d=\frac{v^2-v_0^2}{2 a}=\frac{(8.333)^2-(4.167)^2}{2(200)}$
$\begin{aligned} & =13.02 \\ & =13.0 \mathrm{~m}\end{aligned}$
4 – Potential Difference and Changes in Energy
Equipotential Lines
- As a charge moves along an electric field line, work is done by the electrical force. The energy gained or lost by this charge moving in the field is a form of potential energy, and so associated with the electric field is an electric potential, $\mathrm{V}$, which has units of Energy per charge or Joules per Coulomb (also call Volts).
- Since voltage is potential energy per unit charge, voltage increases when going from a negative charge towards a positive charge. (The kinetic energy of a positive charge would increase when going from a higher potential to a lower potential.)
- A surface along which the potential is constant is called an Equipotential. On a piece of paper, the equipotential is represented by a line on which the voltage is constant.
Topographical Maps:
- Since gravitational potential energy depends on height, lines of constant height would be gravitational equipotentials. A map of such lines is called a topographical map. Typically, a topographical map shows equally spaced lines of constant elevation.
- Where the lines are most closely spaced the elevation is changing most sharply, in other words the terrain is steed.
Potential Difference
We sometimes want to determine the electric potential between two points. This is known as the potential difference.
For example, given two points $\mathrm{A}$ and $\mathrm{B}$, the potential difference between $\mathrm{A}$ and $\mathrm{B}$ is:
$\Delta V=V_B-V_A$
NOTE: When we talk about potential at a point we are talking about the potential difference between that point and infinity, where the potential at infinity is ZERO.
Example:
What is the potential difference between points A and B due to the charge shown?
▶️Answer/Explanation
$\Delta V=V_B-V_A$
$=\frac{k_q}{r_B}-\frac{k_q}{r_A}=k_q\left(\frac{1}{v_B}-\frac{1}{r_A}\right)$
$=\left(9 \times 10^9\right)\left(8 \times 10^{-6}\right)\left(\frac{1}{0.5}-\frac{1}{1.0}\right)$
$=72000 \mathrm{~V}$
Example:
How much work would you have to do to move a proton from point $A$ to point $B$ ?
▶️Answer/Explanation
$W=\Delta E_p=\Delta V_q=(+72000)\left(+1.6 \times 10^{-14}\right)=+1.15 \times 10^{-14} \mathrm{~J}$
Changes in Energy
A $4.0 \times 10^{-9} \mathrm{C}$ charge of mass $2.4 \times 10^{-21} \mathrm{~kg}$, is initially located at point $\mathbf{A}, 3.0 \mathrm{~m}$ from a stationary $6.0 \times 10^{-8} \mathrm{C}$ charge.
Example:
- a) How much work is required, by an external agent, to move the $4.0 \times 10^{-9} \mathrm{C}$ charge to a point $\mathbf{B}, 0.50 \mathrm{~m}$ from the stationary charge?
▶️Answer/Explanation
$\Delta E_p=E_{pt}-E_{p i}=\frac{k_{q_1 q_2}}{r_f}-\frac{k_{q_1 q_2}}{r_i}$
$=k_{q_1} q_2\left(\frac{1}{r_f}-\frac{1}{r_i}\right)$
$=\left(9 \times 10^9\right)\left(4 \times 10^{-9}\right)\left(6 \times 10^{-8}\right)\left(\frac{1}{0.5}-\frac{1}{3.0}\right)$
$=3.6 \times 10^{-6} \mathrm{~J}$
Example:
b) If the $4.0 \times 10^{-9} \mathrm{C}$ charge is now released from point $\mathbf{B}$, what will be its velocity when it passes back through point $\mathbf{A}$ ?
▶️Answer/Explanation
$\Delta E_k=-\Delta E_p=-\left(-3.6 \times 10^{-6} \mathrm{~J}=3.6 \times 10^{-6} \mathrm{~J}\right.$
$\Delta E_k=E_{kf}-E_{ki}$
$E_k=\frac{1}{2} m v^2$
$V=\sqrt{\frac{2 E_k}{m}}$
$\sqrt{\frac{2\left(3.6 \times 10^{-6}\right)}{2.4 \times 10^{-21}}}$
$=5.5 \times 10^7 \mathrm{mls}$
6 – Freefall
- In the absence of air friction…….all objects accelerate at the same rate
- Near Earth’s surface the acceleration is approximately
$g=-9.8 \mathrm{~m} / \mathrm{s}^2$
Example:
A student drops their homework down a wishingwell. After 2.4 s it hits the water at the bottom. How deep is the well?
▶️Answer/Explanation
$V=$
$V_0=$ 0
$a=$ $9.8m/s^{2}$
$d=$
$t=$ 2.4sec
$d=V_0 t+\frac{1}{2} a t^2$
$d=\frac{1}{2} a t^2$
$=\frac{1}{2}(-9.8)(24)^2$
$\begin{aligned} & =-28.22 \mathrm{~m} \\ & =-28 \mathrm{~m}\end{aligned}$
Example:
A football is kicked straight up in the air at $15 \mathrm{~m} / \mathrm{s}$.
a) How high does it go?
▶️Answer/Explanation
$\begin{aligned} & V=0 \\ & V_0=15 \mathrm{~m} / \mathrm{s} \\ & a=-9.8 \mathrm{~m} / \mathrm{s}^2 \\ & d=? \\ & t=\end{aligned}$
$V^2=V_0^2+2 a d$
$0=V_0^2+2 a d$
$-v_0{ }^2-v_0{ }^2$
$\frac{-V_0^2}{2a}=\frac{2 a d}{2 a}$
$d=\frac{-V_0^2}{2 a}$
$d=\frac{-(15)^2}{2(-98)}$
$=11.48$
$11m$
Example:
b) What is its total hangtime?
▶️Answer/Explanation
$\begin{aligned} & V=V_0+a t \\ & -V_0-V_0\end{aligned}$
$\frac{V-V_0}{a}=\frac{a t}{a}$
$t=\frac{V-V_0}{a}$
$=\frac{0+15}{+9.8}$
$=1.531 \mathrm{~s}$
$t_{total \text { } }=2 \times t_{\frac{1}{2}}=2(1.531)=3.06=3.1 \mathrm{s}$
Example:
A student stands on the edge of a $45.0 \mathrm{~m}$ high cliff. They throw their physics homework straight up in the air at $12.0 \mathrm{~m} / \mathrm{s}$.
a. How long does it take to come back down to the same height as the student?
▶️Answer/Explanation
$\begin{aligned} & V=-12.0 \mathrm{m/s} \\ & V_0=12.0 \mathrm{m/s} \\ & a=-9.8 \mathrm{~m} / \mathrm{s}^2 \\ & d=0 \mathrm{~m} \\ & t=\end{aligned}$
$\begin{aligned} & V=V_0+a t \\ & V-V_0=a t \\ & \frac{V-V_0}{a}=t\end{aligned}$
$t=\frac{V-v_0}{a}=\frac{(-12.0)-(12.0)}{-9.8}$
$\begin{aligned} & =2.449 \mathrm{~s} \\ & =2.45 \mathrm{~s}\end{aligned}$
Example:
b. If it falls all the way to the bottom of the cliff, how fast is it traveling when it hits the ground?
▶️Answer/Explanation
$\begin{aligned} & v=? \\ & v_0=12.0 \mathrm{~m} / \mathrm{s} \\ & a=-9.8 \mathrm{~m} / \mathrm{s}^2 \\ & d=-45.0 \mathrm{~m} \\ & t=\end{aligned}$
$\begin{aligned} \sqrt{V^2} & =\sqrt{V_0^2+2 a d} \\ V & =\sqrt{V_0^2+2 a d} \\ & =\sqrt{(12.0)^2+2(+9.8)(+45.0)}\end{aligned}$
$\begin{aligned} & = \pm 32.03 \mathrm{mls} \\ & =-32.0 \mathrm{~m} / \mathrm{s}\end{aligned}$
chepter 3 Force: 4.1
Force: Any push or pull
The units of force are: Newtons (N)
There are four fundamental forces that make up all of the forces in the universe:
1) Gravity
2) Electoromagndic
3) Strong Nuclear Force
4) Weak Nuclear Fore
Force of Gravity: attraction between matter
Mass (kg) total amount of matter in an object.
Weight (N): gravitation al force between two objects.
Mass is Constant throughout the universe but
weight changes depending on where you are.
The formula for force of gravity is:
Force of gravity $Fg=m g$
Where:
- $m=$ mass $\left(K_y\right)$
$\mathrm{g}=$ acceleration due to gravity $=9.8 \mathrm{~m} / \mathrm{s}^2$ (on Earth)
g varies depending on… the size of the planet and how close you are to it.
For Example:
- On Earth at sea level, $g=9.8{m}/ \mathrm{s}^2$
- On the moon, $g=1.6 \mathrm{~m} / \mathrm{s}^2$
- On Jupiter, $g=24.5 \mathrm{~m} / \mathrm{s}^2$
- On the sun, $g=274 \mathrm{~m} / \mathrm{s}^2$
Determine your weight on Earth, the moon and Jupiter (in Newtons)
Your Mass: $100 \mathrm{~kg}(1 \mathrm{~kg}=2.2 \mathrm{lbs})$
Weight on Earth:
$
\begin{aligned}
F_g & =m g \\
& =(100)(9.8)=980 \mathrm{~N}
\end{aligned}
$
Weight on the Moon:
$
F_g=(100)(1.6)=160 \mathrm{~N}
$
Weight on Jupiter:
$
F_g=(100)(24.5)=2450 \mathrm{~N}
$
4.2
Newton’s $1^{\text {st }}$ Law:
An object in motion will..stay in motion
and an object at rest will… Stay at rest
unless…an un balanced force acis upon it
This is also referred to as the Law of Inertia.
Inertia: the tendency for matter to keep dong what it is doing $\rightarrow$ MASS
Imagine that you are racing around a track on a go-kart. List three times when you notice your inertia.
1) Starting up
$\left.\begin{array}{l}\text { 2) Stoping } \\ \text { 3) Tuming }\end{array}\right\}$ acceleration
Another way of thinking of Newton’s $1^{\text {st }}$ Law is that if there is no net force on an object then it will stay at a constant velocity.
If it is not moving then it has a constant velocity of zero!!!
Example
Imagine a book sitting on a table. There is a force of gravity pulling down on the book, but there is also a supporting (normal) force pushing up on the book.
▶️Answer/Explanation
$=$ Normal force
= supporting force
$=$ always perpendicular to the surf ice.
blanced forces
no acceleration
Example
If I drop the book from $2 \mathrm{~m}$, there is only a downwards, gravitational force acting on it. Now that the forces on it are unbalanced, what does the book do?
▶️Answer/Explanation
unbalanced fores
ACCELERATES