PHYS 218 Chapter 2

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Motion In General

• kinematics: how objects move as a function of time
• dynamics: why objects move the way they do

Motion in a Single Direction

Plot position ($x$) vs. time ($t$)

Velocity

VECTOR Displacement from a point in a certain amount of time (if you end up where you started, your velocity is 0)

Acceleration

VECTOR change in velocity = derivative of velocity

Summary

• $x(t)={\frac {dv}{dt}}={\frac {d^{2}a}{dt^{2}}}$
• $\int {x(t)\ dt}=v(t)={\frac {da}{dt}}$
• $\int {\int {x(t)\ dt}\ dt}=\int {v(t)\ dt}=a(t)$

Derived Formulas

1. $x-x_{0}=v_{0}t+{\tfrac {1}{2}}at^{2}$
2. $v=v_{0}+at$
3. $v^{2}={v_{0}}^{2}+2a\left(x-x_{0}\right)$

Solving Constant Acceleration Problems

1. Read the problem. Write the possible variables for each ($x-x_{0}$, $v$, $v_{0}$, $a$, and $t$)
2. Reread the problem and identify all given values. Convert directly to SI units
3. If only one object, then count number of unknowns (more than 2 go to next step).
   If more than one object, determine what variable they have in common
4. Use substitution to get rid of some variables (e.g. $v^{2}=v_{0}^{2}+2a\left(x-x_{0}\right)$)

Free Fall

Acceleration due to gravity:

$g=9.8m/s^{2}\quad a=-g$

Example 1

A ball is thrown upwards from the top of the building with an initial velocity $v_{0}$. What is the speed of the ball when it hits the ground?

Given:

• $x-x_{0}=-h$
• $v_{0}=v_{0}$
• $a=-g=-9.8$

We're not interested in time, so we'll use the third derived equation:

$v={\sqrt {{v_{0}}^{2}+2(-g)(-h)}}={\sqrt {{v_{0}}^{2}+2hg}}$

Example 2

From the previous example, what is the maximum height of the ball from the top of the building?

Given:

• $v=0$
• $v_{0}=v_{0}$
• $a=-g=-9.8$

Using derived equation 3:

${\begin{aligned}0&={v_{0}}^{2}+2(-g)(x-x_{0})\\-{v_{0}}^{2}&=-2g(x-x_{0})\\{\frac {{v_{0}}^{2}}{2g}}&=x-x_{0}\end{aligned}}$

Example 3

A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by $x(t)=bt^{2}-ct^{3}$ where $b$ = 2.4 m/s² and $c$ = 0.12 m/s².

Calculate:

1. instantaneous velocity of the car at t = {0, 5, 10} seconds
2. How long after starting from rest is the car again at rest

${\begin{aligned}v(t)&={\frac {dx}{dt}}=2(2.4)t-3(0.12)t^{2}\\v(\{0,5,10\})&=\{0,15,12\}\\2(2.4)t-3(0.12)t^{2}&=0\\t(4.8-0.36t)&=0\\t&=0,13.{\bar {3}}\end{aligned}}$

Dr. Ricardo Eusebi

Monday, January 31, 2011

Equations of motion

For constant acceleration,

1. ${\overrightarrow {x}}(t)={\overrightarrow {x}}_{0}+{\overrightarrow {v}}_{0}t+{\frac {1}{2}}{\overrightarrow {a}}t^{2}$

Sample Problem 1

A cannon 80m from the ground tilted up at 30° fires a cannon ball at 60m/s. Find the following:

1. Height as a function of time
2. Position in $x$ as a function of time
3. Velocity as a function of time
4. Maximum height, distance of height, and time ($x_{m}$, $y_{m}$, $t_{m}$)
5. Max range ($x_{g}$)
6. Velocity at the time of impact ($t_{g}$), speed as well

Assume, for this problem, that $g=10{\mbox{m/s}}$

1 & 2: height and position as functions of time

${\begin{cases}y(t)=y_{0}+v_{0y}t+{\frac {1}{2}}a_{y}t^{2}\\y(t)=y_{0}+v_{0y}t+{\frac {1}{2}}a_{y}t^{2}\end{cases}}$

Plug in known values:

• $y_{0}=80m$
• $x_{0}=0$
• $v_{0y}=60\sin(30^{\circ })$
• $v_{0x}=60\cos(30^{\circ })$

$={\begin{cases}y(t)=80+30t-5t^{2}\\x(t)=60\cos(30^{\circ })t\end{cases}}$

3: Velocity as function of time

Derivatives of position and height components

${\overrightarrow {v}}(t)={\begin{cases}{\frac {dy}{dt}}=30-10t\\{\frac {dx}{dt}}=60\cos(30^{\circ })\end{cases}}$

4: Max height, reached at time and position

To find time, plug in 0 for $v_{y}(t)$ because at max height, vertical component of velocity is 0:

${\begin{aligned}v_{y}(t=t_{m})&=30-10t_{m}=0\\t_{m}&=3\ {\mbox{seconds}}\end{aligned}}$

Plug 3 seconds back into $y(t)$ to solve for max height:

${\begin{aligned}y_{m}=y(3)&=80+30(3)-5(3)^{2}\\&=125\ {\mbox{meters}}\end{aligned}}$

Plug 3 seconds back into equation $x(t)$ to solve for position:

${\begin{aligned}x_{m}=x(3)&=60\cos(30^{\circ })(3)\\&=180\cos(30^{\circ })\ {\mbox{meters}}\end{aligned}}$

5: Range/ Point of impact

Set $y(t)$ equal to 0 since that's where function crosses $x$-axis

${\begin{aligned}y(t=t_{g})&=80+30t_{g}-5{t_{g}}^{2}\\t_{g}&={\frac {-30\pm {\sqrt {900-4(-5)(80)}}}{2(-5)}}&={\cancel {-2}},8\ {\mbox{seconds}}\end{aligned}}$

(−2 is a non-physical solution)

Plug time into position equation to get position of impact:

${\begin{aligned}x(8)&=60\cos(30^{\circ })(8)\\&=480\cos(30^{\circ })\end{aligned}}$

6: Velocity and speed at impact

${\overrightarrow {v}}={\begin{cases}v_{x}(t_{g})&=60\cos(30^{\circ })\ {\mbox{m/s}}\\v_{y}(t_{g})&=-50\ {\mbox{m/s}}\end{cases}}$

Speed is magnitude of velocity:

$S=\left|{\overrightarrow {v}}\right|={\sqrt {{v_{x}}^{2}+{v_{y}}^{2}}}=\ldots =72.11\ {\mbox{m/s}}$

Sample Problem 2

A balloonist ascending at 5m/s drops a sand bag from 40m

1. What is the position and velocity at t=0.25
2. What is the position and velocity at t=1
3. How many seconds after release will bag hit the ground
4. What magnitude of velocity (speed) does it hit the ground

$y(t)=40+5t-4.9t^{2}$

1. $y(0.25)=\ldots$
2. $y(1)=\ldots$
3. $y(t_{g})=40+5t-4.9t^{2}=0\ \ldots$
4. $y'(t_{g})=5-9.8t$

Summary

For non-constant acceleration:

1. $v(t)=v_{0}+\int _{0}^{t}a\ dt$
2. $x(t)=x_{0}+\int _{0}^{t}v(t)\ dt$

For constant acceleration:

1. $v(t)=v_{0}+at$
2. $x(t)=x_{0}+v_{0}t+{\frac {1}{2}}at^{2}$
3. $v^{2}(t)={v_{0}}^{2}+2a(x(t)-x_{0})$ (derived)
4. $x(t)=x_{)}+\left(frac{v_{0}+v(t)}{2}\right)t$ (derived)

Derived Example

An object is thrown straight up at $v_{0}$ m/s. What is the velocity 20 m up?

• $x-x_{0}=20\ {\mbox{m}}$

$v(t)^{2}={v_{0}}^{2}-2g(20)$

Problem 2.77

A certain volcano on earth can eject rocks vertically to a maximum height $H$.

1. How high (in terms of H) would these rocks go if a volcano on Mars ejected them with the same initial velocity? The acceleration due to gravity on Mars is 3.71 m/s², and you can neglect air resistance on both planets.
2. If the rocks are in the air for a time $T$ on earth, for how long (in terms of $T$) will they be in the air on Mars?

Not given $v_{0}$, so let's set up pictures and problems:

$y(t)=0+v_{0}t-{\frac {g}{2}}t^{2}$

Take derivative for velocity:

$v(t)=v_{0}-gt$

The max height is where $v(t)$ is 0:

$0=v_{0}-gt_{H}\longrightarrow t_{H}={\frac {v_{0}}{g}}$

Plug into original position equation:

$H=y(t_{H})={\frac {{v_{0}}^{2}}{g}}-{\frac {g}{2}}{\frac {{v_{0}}^{2}}{g^{2}}}={\frac {{v_{0}}^{2}}{2g}}$

$H_{M}=y(t_{H})={\frac {{v_{0}}^{2}}{2g_{M}}}$

Find ratio between:

${\frac {H_{M}}{H}}={\frac {{v_{0}}^{2}}{2g_{M}}}{\frac {2g}{{v_{0}}^{2}}}={\frac {g}{g_{M}}}\approx 2.64$