federica bianco
astro | data science | data for good
PHYS 207.013
Chapter 2
1D motion
Instructor: Dr. Bianco
TAs: Joey Betz; Lily Padlow
University of Delaware - Spring 2021
MATH REVIEW: quadratic equations
ax^2 + bx + c = 0
"quadratic" because the highest order is
x^2
it is the equation of a parabola
the "roots" (solutions) are 2 and are given by the quadratic formula
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}
x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}
MATH REVIEW: quadratic equations
MATH REVIEW: quadratic equations
scribbled notes from class... sorry about my poor handwriting...
MATH REVIEW: quadratic equations
ax^2 + bx + c = 0
"quadratic" because the highest order is
x^2
it is the equation of a parabola
the "roots" (solutions) are 2 and are given by the quadratic formula
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}
x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}
coefficients
MATH REVIEW: quadratic equations
ax^2 + bx + c = 0
"quadratic" because the highest order is
x^2
it is the equation of a parabola
the "roots" (solutions) are 2 and are given by the quadratic formula
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}
x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}
second
order
first
order
zeroth
order
H&R CH2 1D motion - velocity - acceleration
motion in 1D
1. coordinates
to define quantities in the 3D space we need a frame of reference: we choose the
origin and direction of an axis
0
positive
negative
notation+conventions
H&R CH2 1D motion - velocity - acceleration
motion in 1D
0
positive
negative
x(t)=2
x(0)=-1
x (t)
2. position
is the position of my object of interest at time t
H&R CH2 1D motion - velocity - acceleration
motion in 1D
notation+conventions
sign: once I chose the positive direction of my axis the sign will fall accordingly.
: greek letter Delta (capital) indicates the difference between two quantities
0
\Delta
\Delta x = x(t) - x(0)
3. displacement
x(t)=2
x(0)=-1
\Delta x = x_\mathrm{final} - x_\mathrm{initial} = x_1-x_0
positive
negative
H&R CH2 1D motion - velocity - acceleration
motion in 1D
0
\Delta x = 2 - (-1) = 3
3. displacement
\Delta x = x_\mathrm{final} - x_\mathrm{initial} = x_1-x_0
x(t)=2
x(0)=-1
I can define it as a vector quantity: it has a magnitude and a direction (along my 1D axis for now)
positive
negative
H&R CH2 1D motion - velocity - acceleration
motion in 1D
We often plot the 1-D motion axis vertically and the time horizontally
0
time
position
4. position over time
x(t=2)=2
x(0)=-1
positive
negative
H&R CH2 1D motion - velocity - acceleration
motion in 1D
\bar{v} = \frac{\Delta x}{\Delta{t}}
this is a vector quantity
H&R CH2 1D motion - velocity - acceleration
0
t
x
\Delta(x)
\Delta(t)
AVERAGE VELOCITY
5.a average velocity
H&R CH2 1D motion - velocity - acceleration
motion in 1D
5.a average velocity
\mathrm{speed :} \frac{\mathrm{total~distance~traveled}}{\Delta t}
H&R CH2 1D motion - velocity - acceleration
0
t
x
\Delta(x)
\Delta(t)
5.a average velocity
\bar{v} = \frac{\Delta x}{\Delta{t}}
AVERAGE VELOCITY
H&R CH2 1D motion - velocity - acceleration
motion in 1D
\bar{v} = \frac{\Delta x}{\Delta{t}}
\vec{v}(t) = \frac{d \vec{x}}{d{t}} =\lim_{\Delta t\to 0} \frac{\Delta \vec{x}}{\Delta{t}}
x
t
AVERAGE VELOCITY
INSTANTANEOUS VELOCITY
5.b instantaneous velocity
H&R CH2 1D motion - velocity - acceleration
motion in 1D
\bar{a} = \frac{\Delta \vec{v}}{\Delta{t}}
H&R CH2 1D motion - velocity - acceleration
AVERAGE VELOCITY
6.a average acceleration
t
0
t
v
\Delta(v)
\Delta(t)
v(t=2)=1
v(0)=-1
H&R CH2 1D motion - velocity - acceleration
motion in 1D
\bar{a} = \frac{\Delta \vec{v}}{\Delta{t}}
H&R CH2 1D motion - velocity - acceleration
AVERAGE VELOCITY
6.b instantaneous acceleration
\vec{a}(t) = \frac{d \vec{v}}{d{t}} =\lim_{\Delta t \to 0} \frac{\Delta \vec{v}}{\Delta{t}} = \frac{d^2 \vec{v}}{d{t^2}}
INSTANTANEOUS acceleration
example of motion graph:
position
velocity
acceleration
H&R CH2 1D motion - velocity - acceleration
H&R CH2 1D motion - velocity - acceleration
KEY POINTS:
- displacement is the distance between final and initial position
- average velocity is displacement over time
- average acceleration is change of velocity over time
- instantaneous velocity is the derivative of displacement in time
- instantaneous acceleration is the derivative of velocity in time
H&R CH2 1D motion - velocity - acceleration
motion in 1D
H&R CH2 1D motion - velocity - acceleration
Equations of motion in 1D
you do not need to memorize them necessarily but you should be extremely familiar with them and know how to use each one.
When you try and remember them on the fly, think about the dimensional analysis.
To equate to displacement, velocity needs to be multipled by time.
To equate displacement acceleration has to be multiplied by time-squared
In a few weeks, you should have solved enough problems that you will have them memorized without even trying!
H&R CH2 1D motion - velocity - acceleration
H&R CH2 1D motion - velocity - acceleration
- motion in 1D can be described by a quadratic equation
KEY POINTS:
- in the equation of motion the first order coefficient is velocity
- in the equation of motion the second order coefficient is acceleration
- displacement is the distance between final and initial position
- average velocity is displacement over time
- average acceleration is change of velocity over time
- instantaneous velocity is the derivative of displacement in time
- instantaneous acceleration is the derivative of velocity in time
{
{
{
H&R CH2 1D motion - velocity - acceleration
free-fall
H&R CH2 1D motion - velocity - acceleration
An exemplary case of constant acceleration
a = g = -9.8 m/s
| Mass [Me] | g [m/s] | |
|---|---|---|
| Venus | 0.05539 | 8.83 |
| Mercury | 0.815 | 3.61 |
| Earth | 1 | 9.8 |
| Mars | 0.1075 | 3.38 |
| Jupiter | 317.8 | 26.0 |
| Saturn | 5.2 | 11.2 |
| Uranus | 14.6 | 10.5 |
| Neptune | 17.2 | 13.3 |
H&R CH2 1D motion - velocity - acceleration
free-fall
H&R CH2 1D motion - velocity - acceleration
a = g = -9.8 m/s
Two identical objects are dropped from the Tower of Pisa 1 second apart. What happens to the distance between the 2 balls during the
a) Remain the same.
b) Decrease.
c) Increase.
d) First increase, then decrease.
e) First decrease, then increase.
discuss for 8 min:
a) which is the right anwer?
b) which equation would you use to show it?
H&R CH2 1D motion - velocity - acceleration
free-fall
H&R CH2 1D motion - velocity - acceleration
Two identical objects are dropped from the Tower of Pisa 1 second apart. What happens to the distance between the 2 balls during the
H&R CH2 1D motion - velocity - acceleration
free-fall
H&R CH2 1D motion - velocity - acceleration
Two identical objects are dropped from the Tower of Pisa 1 second apart. What happens to the distance between the 2 balls during the
free-fall
Two identical objects are dropped from the Tower of Pisa 1 second apart. What happens to the distance between the 2 balls during the
Because the distance is propotional to time
the distance increases linearly with time
\Delta x = \frac{1}{2} g t
HOW TO SOLVE 1D MOTION PROBLEMS
STEP 1. Put together all the information you have
STEP 2. Put the info in a frame of reference
STEP 3. Identify the unknown quantity
STEP 4. Choose the equation with gives you the unknown you need
A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.
(a) How long does the package take
HOW TO SOLVE 1D MOTION PROBLEMS
STEP 1. Put together all the information you have
STEP 2. Put the info in a frame of reference
STEP 3. Identify the unknown quantity
STEP 4. Choose the equation with gives you the unknown you need
A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.
(a) How long does the package take
HOW TO SOLVE 1D MOTION PROBLEMS
v_0 = 12 \frac{m}{s} ~\mathrm{upward}
A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.
(a) How long does the package take
STEP 1. Put together all the information you have
STEP 2. Put the info in a frame of reference
STEP 3. Identify the unknown quantity
STEP 4. Choose the equation with gives you the unknown you need
a = g = 9.8 \frac{m}{s^2} ~\mathrm{downward}
x_0 = 0
x = 80m
HOW TO SOLVE 1D MOTION PROBLEMS
v_0 = -12 \frac{m}{s}
A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.
(a) How long does the package take
STEP 1. Put together all the information you have
STEP 2. Put the info in a frame of reference
STEP 3. Identify the unknown quantity
STEP 4. Choose the equation with gives you the unknown you need
a = g = 9.8 \frac{m}{s^2}
x_0=0
x=80m
x_0 = 0
x = 80m
HOW TO SOLVE 1D MOTION PROBLEMS
x_0 = 0
v_0 = -12 \frac{m}{s}
A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.
(a) How long does the package take
STEP 1. Put together all the information you have
STEP 2. Put the info in a frame of reference
STEP 3. Identify the unknown quantity
STEP 4. Choose the equation with gives you the unknown you need
x_0=0
x=80m
\mathbf{t = ?}
x = 80m
a = g = 9.8 \frac{m}{s^2}
HOW TO SOLVE 1D MOTION PROBLEMS
x_0 = 0
v_0 = -12 \frac{m}{s}
A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.
(a) How long does the package take
STEP 1. Put together all the information you have
STEP 2. Put the info in a frame of reference
STEP 3. Identify the unknown quantity
STEP 4. Choose the equation with gives you the unknown you need
x_0=0
x=80m
\mathbf{t = ?}
x = 80m
a = g = 9.8 \frac{m}{s^2}
HOW TO SOLVE 1D MOTION PROBLEMS
x_0 = 0
v_0 = -12 \frac{m}{s}
A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.
(a) How long does the package take
x_0=0
x=80m
\mathbf{t = ?}
x = 80m
STEP 1. Put together all the information you have
STEP 2. Put the info in a frame of reference
STEP 3. Identify the unknown quantity
STEP 4. Choose the equation with gives you the unknown you need
a = g = 9.8 \frac{m}{s^2}
HOW TO SOLVE 1D MOTION PROBLEMS
x_0 = 0
v_0 = -12 \frac{m}{s}
A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.
(a) How long does the package take
STEP 1. Put together all the information you have
STEP 2. Put the info in a frame of reference
STEP 3. Identify the unknown quantity
STEP 4. Choose the equation with gives you the unknown you need
x_0=0
x=80m
\mathbf{t = ?}
x = 80m
a = g = 9.8 \frac{m}{s^2}
HOW TO SOLVE 1D MOTION PROBLEMS
x_0 = 0
v_0 = -12 \frac{m}{s}
A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.
(a) How long does the package take
a = g = 9.8 \frac{m}{s^2}
x_0=0
x=80m
\mathbf{t = ?}
x = 80m
x - x_0 = vt + \frac{1}{2} a t^2
80 - 0 = - 12t + \frac{1}{2} 9.8 t^2
HOW TO SOLVE 1D MOTION PROBLEMS
x_0 = 0
v_0 = -12 \frac{m}{s}
A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.
(a) How long does the package take
a = g = 9.8 \frac{m}{s^2}
x_0=0
x=80m
\mathbf{t = ?}
x = 80m
x - x_0 = vt + \frac{1}{2} a t^2
a t^2 + b t + c = 0
write it as a quadratic equation in t
\frac{1}{2} 9.8 t^2 - 12t - 80 = 0
80 - 0 = - 12t + \frac{1}{2} 9.8 t^2
HOW TO SOLVE 1D MOTION PROBLEMS
x_0 = 0
v_0 = -12 \frac{m}{s}
A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.
(a) How long does the package take
a = g = 9.8 \frac{m}{s^2}
x_0=0
x=80m
\mathbf{t = ?}
x = 80m
x - x_0 = vt + \frac{1}{2} a t^2
a t^2 + b t + c = 0
write it as a quadratic equation in t
\frac{1}{2} 9.8 t^2 - 12t - 80 = 0
80 - 0 = - 12t + \frac{1}{2} 9.8 t^2
t = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}
HOW TO SOLVE 1D MOTION PROBLEMS
x_0 = 0
v_0 = -12 \frac{m}{s}
A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.
(a) How long does the package take
a = g = 9.8 \frac{m}{s^2}
x_0=0
x=80m
\mathbf{t = ?}
x = 80m
x - x_0 = vt + \frac{1}{2} a t^2
a t^2 + b t + c = 0
write it as a quadratic equation in t
\frac{1}{2} 9.8 t^2 - 12t - 80 = 0
80 - 0 = - 12t + \frac{1}{2} 9.8 t^2
t = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}
=\frac{12 \pm \sqrt{144 + 4\frac{9.8}{2}80}}{2\frac{9.8}{2}}
=\frac{12 \pm \sqrt{144 + 4\frac{9.8}{2}80}}{2\frac{9.8}{2}}
5.4s
-3.0s
HOW TO SOLVE 1D MOTION PROBLEMS
x_0 = 0
v_0 = -12 \frac{m}{s}
A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.
(a) How long does the package take
a = g = 9.8 \frac{m}{s^2}
x_0=0
x=80m
\mathbf{t = ?}
x = 80m
x - x_0 = vt + \frac{1}{2} a t^2
a t^2 + b t + c = 0
write it as a quadratic equation in t
\frac{1}{2} 9.8 t^2 - 12t - 80 = 0
80 - 0 = - 12t + \frac{1}{2} 9.8 t^2
t = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}
=\frac{12 \pm \sqrt{144 + 4\frac{9.8}{2}80}}{2\frac{9.8}{2}}
=\frac{12 \pm \sqrt{144 + 4\frac{9.8}{2}80}}{2\frac{9.8}{2}}
5.4s
-3.0s
References reg space exploration
https://justspacealliance.org/quickreads
phys207 1D motion
By federica bianco
phys207 1D motion
1D motion, coordinates, displacement, velocity, acceleration,
