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## AP®︎/College Physics 1

### Course: AP®︎/College Physics 1 > Unit 12

Lesson 1: AP Physics 1 concept review- AP Physics 1 review of 1D motion
- AP Physics 1 review of 2D motion and vectors
- AP Physics 1 review of Forces and Newton's Laws
- AP Physics 1 review of Centripetal Forces
- AP Physics 1 review of Energy and Work
- AP Physics 1 review of Momentum and Impulse
- AP Physics 1 review of Torque and Angular momentum
- AP Physics 1 review of Waves and Harmonic motion
- AP Physics 1 Review of Charge and Circuits

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# AP Physics 1 review of 1D motion

In this video David rapidly explains all the concepts in 1D motion and also quickly solves a sample problem for each concept. Keep an eye on the side scroll see how far along you've made it in the review video. Created by David SantoPietro.

## Want to join the conversation?

- How can I get this document?(12 votes)
- Here's a dropbox link to download them. You can click the x to avoid having to register for dropbox. https://www.dropbox.com/s/8satagub57ldq5b/1D%20motion%20AP%201%20concept%20sheets.pdf?dl=0

Good luck on the exam!(47 votes)

- hey what does a negative displacement interpret...what does negative displacement mean...for how could my speed be negative...is it not speed scalar...someone pls help me with the concept(3 votes)
- Displacement is relative, which means that negative displacement will be the opposite of what you define as positive. For example, if walking forward is a positive displacement, then going backwards is a negative displacement. Speed just measures how fast you are going, regardless of direction. Velocity, however, has a direction can be negative(saw my mistake in comments).(4 votes)

- Has anyone taken the exam this year already?

I would love some general concepts to focus on and any tips that could help.(2 votes)- Hey,

I took the AP Physics 1 Exam this year. My main advice would be to really focus on understanding the theory behind the content (as opposed to just plug/chug) - this conceptual framework behind the exam makes it really different from other physics exams. Other resources include FlippingPhysics, APlusPhysics, and of course this website. Do tons of practice questions to get used to the style of questions on the Exam, which contains a LOT of reading/writing; also use the official AP 1 formula sheet when doing practice problems, so you are prepared for exam day.

Good Luck,

NP(7 votes)

- how would you find the initial velocity for an object released from a height if only the height and distance are given?(3 votes)
- The object is merely released from a height? Why, then, the initial velocity must equal zero! :)(5 votes)

- Why is acceleration velocity over time and not distance over time?(2 votes)
- By definition.

velocity is rate of change of displacement.

Acceleration is rate of change of velocity.(5 votes)

- I have a book about physics that I use to study. It doesn't explain it the same way as David SantoPietro. Can anyone recommend a book to use that explains 1D motion the same way as him? Also, does Kahn Academy have textbooks, so when I travel on a plane, I can do Kahn Academy at that time?(3 votes)
- While I can't recommend you to a textbook that teaches similarly to David, I can tell you that a digital textbook company called Kno is integrating Khan Academy into their etextbooks. That's probably your best bet.(2 votes)

- At3:00how did you get 4 meters every 2 seconds instead of 2 meters every 2 seconds? On the example problem, there are arrows explaining how every two seconds 4 meters has passed, but can you explain why it is 4 meters and not 2?(2 votes)
- So the time on the x axis is shown in increments of 2 seconds right? And the graph shows that every 2 seconds the curve travels 4 meters since that is the max height (amplitude of the graph). Therefore, since the curve always travels 4 meters every two seconds (regardless of direction), it must travel 2 meters every second and therefore the average speed is 2 meters per second.(4 votes)

- Under the section "Freely Falling Object," how is the answer t = radical 2h / g? I don't understand how that answer was gotten to.(2 votes)
- The equation x=x0+v0t+1/2at^2.

x0=0 and v0=0, so you are left with x=(1/2)at^2.

The distance the textbook has fallen is H, so x=-H. Your equation is now -H=(1/2)at^2.

When you have a free falling object, the acceleration is always -9.8m/s^2 (gravity). If g=9.8m/s^2, you can substitute -g in for 'a' to get -H=(1/2)(-g)t^2.

Then you solve for time.

The negatives on both sides of the equation cancel out => H=(1/2)gt^2

Multiply both sides by 2 => 2H=gt^2

Divide both sides by g => 2H/g=t^2

Then square root both sides, and you should have the answer!(4 votes)

- Will I miss something if I skip the lectures from 1D motion to Quantum Physics? By just watching all these review videos?(3 votes)
- Where do I find practice for the AP Physics 2 course? I get the theory part but unable to find the practice questions and exercises.(2 votes)

## Video transcript

- [Instructor] What does distance mean? In physics, distance is
the total length traveled for a particular trip. In other words, if an object went forward then backward then forward again, to find the total distance traveled you would take this positive path length, add it to this positive path length, and then add this positive path length, which would give you the
total distance traveled. And that's why a formula for distance can be thought of as the summation of all the individual
positive path lengths. The units for distance are meters. It is not a vector and it's
always positive or zero. It can never be negative. What's an example problem
for distance look like? Let's say you have this graph and the question was what's the distance traveled for the object in the graph between zero and six seconds? So between zero and two seconds the object moved forward four meters. Between two and four seconds the object move backward four meters. And between four and six seconds the object moved backward
another four meters, which means the total distance would be four meters plus four
meters plus four meters, which would give you 12
meters of distance traveled. What does displacement mean? In physics, displacement is
the difference in position. So if an object went forward then backward then forward again, the displacement could be
represented with an arrow that points from the initial position all the way to the final position. And that's why a formula for displacement can be thought of as the final position minus the initial position. This delta here represents the change in, which means final minus initial. The S.I. units for
displacement are meters. It's a vector, which means
that the displacement is leftward or downward. The displacement can be negative. In other words, if you end up further left from where you started, your horizontal displacement's negative. And if you end up further
down from where you started, your vertical displacement's negative. What's an example problem
for displacement look like? If you were given this graph and you were asked to
find the displacement of the object between
zero and six seconds, you wouldn't care that
the object moved forward and backward and changed directions. All that you would focus on is that the initial position was zero. The final position at six
seconds was negative four. That means the displacement
was negative four meters since you ended four meters
behind where you started. What does speed mean? Speed is a way to measure
how fast something's moving. And it's a scaler,
which, for our purposes, is really just another way
of saying not a vector. Speed is defined to be
the distance per time. You could define the average speed as the distance per time
over a large time interval. Or you could define
the instantaneous speed as the rate of distance per time at a particular moment in time. In other words, if an object's
taking a long winding path, you could find the average
speed for the entire trip by taking the total
distance for the entire trip and dividing by the time it
took to travel that distance. Or to determine the instantaneous speed, you could look at an
infinitesimally small distance and divide by the time it took to travel that infinitesimally
small distance. The units for speed are meters per second. It is not a vector. And speed can only be positive or zero. Speed can never be negative. What's an example problem
for speed look like? If you had this graph and you wanted to
determine the average speed between zero and six seconds, you'd have to remember that average speed is the distance per time. The distance between zero and six seconds would be four meters plus four meters plus another four meters would
be 12 meters of distance. And the time it took to travel
that 12 meters of distance was six seconds, which
gives us an average speed of two meters per second. What does velocity mean? Velocity's another way to determine how fast something's moving. But this one's a vector, which means velocity can
be positive or negative. Velocity's defined to be
the displacement per time. You could define the average velocity as the displacement per time
over a large time interval. Or you could define the
instantaneous velocity as the rate of displacement per time at a particular moment in time. So in other words, if an object took a long winding path and you wanted to determine
the average velocity for the entire trip, you would take the displacement
for the entire trip and divide by the time it
took for that displacement. Or if you wanted the
instantaneous velocity at a particular moment, you would take an
infinitesimal displacement at that moment and divide
by the time it took for that displacement. The units of velocity
are meters per second. And it is a vector. That means it can be negative. So if the velocity is
directed leftward or downward, we typically consider the
velocity to be a negative value. What's an example problem
for velocity look like? If you had this graph and
you wanted to determine the average velocity of the object between zero and six seconds, you'd have to remember
that average velocity is the displacement per time. Since the object starts at zero meters and ends at six seconds
at negative four meters, the displacement here is negative four. And the time it took to
travel that displacement was six seconds, which
gives us an average velocity of negative 2/3 meters per second for the time interval
between zero and six. What does acceleration mean? Acceleration is the rate
at which velocity changes. You could change your velocity by speeding up, slowing down, or by changing directions by turning. And it's good to remember that
the acceleration of an object always points in the same direction as the net force on that object. So in other words, if you had a ball that had an initial velocity, vi, to the right and the ball speeds up to a larger v final to the right, there must have acceleration
since the velocity changed. And since acceleration always points in the direction of
the change in velocity, since this object gained
velocity to the right, the acceleration points to the right, which also means there must have been a net force on this ball
to the right as well. The formula for acceleration
is the change in velocity over the time it took for
that velocity to change. Since delta means final minus initial, you could write this as the final velocity minus the initial velocity
over the time it took for that velocity to change. The units of acceleration
are meters per second squared since acceleration's telling
you how many meters per second the velocity is changing by per second. And acceleration is a vector, which means it can be negative. If the acceleration points left or down, we typically consider that acceleration to have a negative value. What's an example problem
for acceleration look like? Let's determine the two cases where the acceleration of
this ball points to the right. So if a ball's moving to the
right and it's slowing down, there's gotta be a net force to the left that slows it down. And if the net force points to the left, the acceleration also points to the left 'cause the net force and the acceleration always point in the same direction. And if a ball is moving to the
right and it's speeding up, there's gotta be a net force to the right that's speeding it up. But that also means the acceleration points to the right as well. If a ball is moving leftward
and it's slowing down, there's gotta be a net force
to the right slowing it down, which means the acceleration
also points to the right. And if a ball is moving
leftward and it's speeding up, there's gotta be a net force to the left speeding the ball up, which means the acceleration
also points to the left. So since acceleration is always pointing in the same direction as the net force, we can see that B and C are the two cases where acceleration points to the right. How do you interpret a
position versus time graph? Well, the value of the graph
is giving you the position x. In other words, at some time, t, if you measure the value of the graph, on the vertical axis
that value is giving you the position at that time, t. Why do we care about
position versus time graphs? One reason we care is that the slope is gonna equal the velocity of the object. So if you can determine the slope, you can find the velocity. How do you find the slope? You find the region that you
want to determine the slope in, and then you take the rise over the run, which is the change in x
over the change in time for that portion of the graph. Since the rise over run here
is displacement per time, the slope is equal to the velocity. Also, you can tell if there's acceleration on a position versus time graph depending on whether
there's curvature or not. Curvature that looks like a smiley face represents positive acceleration. And curvature that
looks like a frowny face represents negative acceleration. Graphs that have constant slope would have constant velocity
and zero acceleration. What would an example problem with position versus
time graphs look like? Let's say we have this graph
and we wanted to determine the instantaneous
velocity at seven seconds. So we locate seven seconds. To find the velocity in this region we'll just find the slope in this region. Since the slope is constant, we can find the slope between
any two points in this region. Choosing six seconds and eight seconds will be the most convenient. So we'll use rise over run. The rise in this case would
be the change in position. The run would be the change in time. Since between six
seconds and eight seconds the graph drops by four meters, the displacement would
be negative four meters. And the time it took for that to happen, i.e., the run, would be two seconds. And since the slope in this region is negative two meters per second, that also equals the
velocity in this region. How do you interpret a
velocity versus time graph? Well, the first thing to know is that the value of the
graph on the vertical axis is giving you the value of the velocity at that particular moment in time. Why do we care about
velocity versus time graphs? One reason we care is that
the value of the slope at any given moment is gonna be equal to the acceleration of
the object at that moment. You find the slope using the definition, which is rise over run. And in this case, the rise
is gonna be the change in v. The run is gonna be the change in t. And since the change in velocity over the change in time is the
definition of acceleration, the slope on a velocity graph
is equal to the acceleration. Also, the area under any
particular section of the graph between two times is gonna
equal the displacement of the object between those two times. So what would an example problem involving velocity versus
time graphs look like? Say you had this graph and
you wanted to determine the acceleration of the
object at two seconds. The acceleration's gonna be equal to the slope at two seconds. So we need to find the
slope in this region. Since the slope is constant, we can find the slope
between any two points between zero and four seconds. I'm just gonna pick zero
seconds and four seconds because they're convenient. And we'll find the slope, which is always the rise over the run. The rise in this case is
the change in velocity. The run is the change in time. The change in velocity between
zero seconds and four seconds would be negative 60 meters per second. And the change in time
between zero and four seconds would be four seconds. This gives a slope of negative
15 meters per second squared, which is also the acceleration
in that time period. And if we want to determine
the displacement of the object between four and six seconds, we can determine the area under the graph between four and six seconds. The area under a rectangle
is gonna be equal to the height times the width. The height in this case is gonna be equal to negative 30. And the width in this case is
gonna be equal to two seconds, which gives us an area
of negative 60 meters, which means the displacement
between four and six seconds was negative 60 meters. How do you interpret an
acceleration versus time graph? Well, the value of the vertical axis is gonna give you the acceleration. So if you read the graph at
a particular moment in time, the value of that vertical axis is giving you the
acceleration of the object at that moment in time. The slope in this case is gonna be equal to the jerk, which really isn't
asked about all that much. But you do need to know
that the change in velocity is gonna be represented by
the area under this graph. So if you can find the
area under the graph between two times, that area is gonna equal
the change in velocity experienced by that object
between those two times. So what would an example problem involving acceleration
versus time graphs look like? Well, let's say an
object started from rest at t equals zero and you want to determine with this acceleration
graph what the velocity was at six seconds. We know that the area is gonna represent the change in velocity. So since I'm going from
zero seconds to six seconds, I want to determine the
area under the graph between zero and six seconds. This first area would be a positive area. This second area would be a negative area. And this third area would
also be a negative area. But note that this first triangular area will cancel with this
second triangular area since one is positive and the
other is equally negative. And the only important area in this case is gonna be this rectangle, which the height is negative 10 and the width is two seconds, which gives us negative 20
meters per second for the area. And that's also gonna equal
the change in velocity. Now since the object started
from rest at t equals zero and it changed its
velocity by negative 20, the final velocity at six seconds is gonna equal negative
20 meters per second. What are the kinematic formulas? The kinematic formulas
are formulas that relate the five kinematic variables. Those variables are displacement, initial velocity, final velocity, acceleration, and time. But these kinematic formulas
only give true relationships if the acceleration is constant. These three kinematic formulas
are given on the AP test, but this last one is not. One way to remember this formula is that the left-hand side is v final plus v initial over two, which is the average of the velocities. And the right-hand side
is displacement per time, which is the definition
of average velocity. So it says that the
average of the velocities is equal to the average velocity. So in other words, during some motion, it'll take a certain time, t, for an object to change its velocity from v initial to v final. And during that time, the object will have a certain acceleration and displacement, which will all be
related by these formulas if that acceleration is constant. What would an example problem involving kinematic formulas look like? Let's say a confused
chipmunk started from rest and sped up with constant
acceleration for three seconds traveling nine meters in the process. We could figure out the
acceleration of the chipmunk using the second kinematic formula. The displacement would be nine meters. The initial velocity would be zero since the chipmunk started from rest. The acceleration is the unknown, and the time was three seconds, which, solving for the acceleration gives us two meters per second squared. We could also solve for the
final speed of the chipmunk using the first kinematic formula. The final speed would be our unknown. The initial speed is still zero. We know the acceleration is
two and time was three seconds, which gives a final velocity
of six meters per second. And the reason we could use these formulas is because we knew the
acceleration was constant. A freely falling or flying object is any object that has been dropped or thrown through the air that only has the force
of gravity acting upon it. This means we'll typically ignore air resistance on these objects
unless otherwise stated. We care about these objects because the acceleration
in the vertical direction for all freely falling and
flying objects near the earth will be negative 9.8
meters per second squared. And since this acceleration
is gonna be constant, we can use the kinematic formulas to describe the motion of freely
flying or falling objects. We just have to plug negative 9.8 in for the vertical acceleration. Watch out for code words. Dropped is code word for
initial velocity is zero. The final velocity at the
maximum height will be zero. And the acceleration due to gravity in the vertical direction
for a freely flying object is always negative 9.8
meters per second squared. But be careful. If you analyze the motion
of an object flying upward, the velocity gets smaller on the way up until the velocity is zero at the top, after which the velocity
gets more and more negative on the way down. However, the acceleration
remains a constant negative 9.8 the entire trip, on the way up, on the way down. And even at the top of the motion, the acceleration in the vertical direction is still negative 9.8
meters per second squared. So what would an example problem involving freely flying objects look like? Let's say a student drops
a book from a height, H, and we want to know how long it takes for the book to hit the ground. But we need to represent it in terms of given quantities and
fundamental constants. That means we're gonna have to solve this problem symbolically. So we still use a kinematic formula, but we'll plug in symbolic
values instead of numbers. So delta y is not gonna be H. It's gonna be negative H
since the book dropped down, which is a negative vertical displacement. The book was dropped. That means the initial velocity was zero. The acceleration due to gravity in the y direction is negative 9.8. But since we're solving
this problem symbolically, we're not gonna use an actual number. We're gonna represent
this as negative little g. What we mean by little g is the magnitude of the acceleration due to gravity, which means little g is positive 9.8. Since the acceleration due
to gravity is negative 9.8, we can represent the
acceleration due to gravity as negative g. Now we can solve for our time. And we get the square
root of two H over g, which is our symbolic answer in terms of given quantities
and fundamental constants.