AP Physics In Class Notes Unit 1
Name:
Position Time Graphs
Slope = rate of change.
The rate of change of position with respect to time is _____________________________.
Whatever characteristics the slope has, the velocity will have as well (and vice versa).
1. (a) Describe the graphs below with the following terms:
constant velocity vs changing velocity
positive vs negative
slow vs fast or speeding up vs slowing down
(b) Draw a corresponding velocity vs. time graph and an acceleration vs. time graph.
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Velocity Time Graphs
Slope give rate of change.
The rate of change of velocity with respect to time is _____________________________.
Whatever characteristics the slope has, the acceleration will have as well (and vice versa).
2.
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4
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5.
Using the velocity time graph below
(a) Determine the acceleration of the particle from 0 to 5 seconds. (ans = 4)
(b) Determine the displacement of the particle 0 to 5 seconds. (ans = 75)
CONCLUSIONS:
 Velocity = ____________ of position vs. time graph

Acceleration = _______________ of velocity vs. time graph

One can find displacement from a velocity vs. time graph by _____________________________.

One can find velocity from an acceleration vs. time graph by _____________________________.
Flow Chart
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6. Draw Qualitative Graphs of x vs. t, v vs. t, and a vs. t for the following incline plane situations.
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Instantaneous Velocity and Acceleration:
How can we find the velocity or the acceleration of an object at a single instant in time?
How can we work backwards to find displacement from velocity or to find velocity from acceleration?
We us CALCULUS!! Derivatives and integrals make this very easy!!!
Flow Chart
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A rock thrown vertically upward from the surface of the moon reaches a height of x(t) = 24t – 0.8t2, where t is
measured in seconds and x is measured in meters. (a) Find the velocity of the rock as a function of time. (b) Find
the acceleration of the rock as a function of time. (c) Determine the instantaneous velocity of the rock at t = 2
seconds. (answer = c = 20.8)
8. Paul has a bought a ticket at an amusement park for on a special roller coaster that moves in a straight line. The
velocity of the car in feet per second is given by v(t) = -3t2 + 240t. (a) Find the position of the roller coaster after t
seconds. Assume x(0) = 0 (b) Find the acceleration of the roller coaster as a function of time. (c) Determine the
instantaneous acceleration at t = 3 seconds. (answer = c = 222)
9. A particle’s velocity as a function of time is given by v(t)=2 + 6t – t2, where t is measured in seconds and v(t) is
measured in m/s. (a) Determine the acceleration as a function of time. (b) Determine the position of the particle as
a function of time. Assume x(0) = 0. (c) Find the acceleration of the particle at t = 5 seconds. (answer = c = - 4)
10. A particle’s position is given by x(t) = 6 t , where t is measured in seconds and x is measured in feet. (a)
Determine the velocity of the particle as a function of time. (b) Determine the acceleration of the particle as a
function of time.
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Please thoroughly show all of your work for these problems. At the beginning of each problem, I want you to
write out the flow chart and circle where you are starting and draw an arrow to where you are heading. Lastly I
want you to write what process you plan to take, either differentiate or integrate.
11.
A cheetah position is given by x(t) =
1
, determine the acceleration of the cheetah as a function of time. Assume
t3
x is measured in meters and t is measured in seconds.
12.
A car’s velocity is given by v(t) = 3t2 – 12t + 9, measured in meters per second. (a) Find an equation for the
position of the car as a function of time. Assume x(0) = 0. (b) Find an equation for the acceleration of the car as
a function of time. (c) Find the car’s instantaneous acceleration at t = 2 seconds. (answer = 0)
13.
A runner’s position is given by –t3 + 9t2 – 24t + 1, where t is measured in seconds and x is measured in meters.
(a) Find an equation for the velocity of the car as a function of time. (b) Find an equation for the acceleration of
the car as a function of time.
14.
John’s car runs out of gas as it goes up a hill. The car rolls to a stop then starts rolling backwards. As it rolls, its
acceleration a(t) measured in feet per seconds squared since the car ran out of gas is given by a(t) = 75 + 32t –
t2. (a) Give an equation for John’s velocity as a function of time. Assume v(0)= 0 (b) Give an equation for the
position as a function of time. Assume x(0) = 0.
15.
In driving in a straight line from New York to Boston, your position function given in miles from New York is
described by the function below. Assume t is the number of hours since the trip began. Find your velocity at time
t = ½ hours. (answer = 96.25)
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Motion with a Constant Acceleration
What does constant acceleration mean?
Let us predict what this might look like when one-dimensional motion under constant acceleration is graphed.
Please show a position, velocity and acceleration versus time graph.
Three kinematic equations that hold true for UAM (uniformly accelerated motion)
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1.
vf = vi + at
2.
xf = xi + vit + ½ at2
3.
vf2 = vi2 + 2a(xf-xi)
A bowler releases a ball with an initial velocity of 3.0 m/s; the ball slows with a constant negative
acceleration of –0.20 m/s2. How far does the ball roll before stopping, and how long does it take to
stop?
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17.
A runner bursts out of the starting block 0.10 s after the gun signals the start of a race. She runs at
constant acceleration for the next 1.9 s of the race. If she has gone 8.0 m after 2.0 s, what are her
acceleration and velocity at this time?
18.
A T-38 training jet has an acceleration of 3.6 m/s2 that lasts 5.0 s during the initial phase of takeoff.
The after burner engines are then turned up to full power for an acceleration of 5.1 m/s2. The speed
needed for takeoff is 164 knots (1 m/s = 1.94 knots). Calculate the length of runway needed and total
time of takeoff.
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Freely Falling Objects
The most important example of constant acceleration is gravity. The acceleration due to gravity is given by the symbol g
and its magnitude is 9.8 m/s2. The motion of an object falling at 9.8 m/s2 near the Earth’s Surface is called free fall.
Galileo was the first to discover that the distance an object fell if the object started from rest is proportional to the
square of the time; equivalently, the speed of a falling object is proportional to the square root of the distance fallen.
Try the experiment of dropping a piece of paper and a golf ball. What happens?
Try crinkling the piece of paper and dropping the golf ball and the paper at the same time. What happens now? Is the
result different? Why?
To study the motion of freely falling objects the following formulae can be examined. Notice that the acceleration, a,
has been replaced by the acceleration due to gravity, g.
1.
vf = vi + gt
2.
xf = xi + vit + ½ gt2
3.
vf2 = vi2 + 2g(xf-xi)
One thing to keep in mind when using these equations is where or not the object is accelerating or decelerating. If the
object is accelerating then g is positive. If the object is decelerating then g is negative.
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19.
How much time has elapsed after a ball has been dropped from rest from a height of 100 m? What is
the velocity when the ball hits the ground?
20.
Calculate the position of the ball 1.0 second and 2.0 seconds after the ball has been dropped from the
100 m height.
21.
A ball thrown straight up with a speed of 10.0 m/s from a third floor window that is located 15.0 m
above the ground. Calculate the maximum height of the ball, the balls velocity when it hits the ground,
and the total time it takes to reach the bottom.
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22. In the Blackhawk landslide in California, a mass of rock and mud fell 460 m down a mountain and then traveled 8 km
across a level plain on a cushion of compressed air. Assume that the mud dropped with the free-fall acceleration due to
gravity and then slid horizontally with constant deceleration.
(a) How long did the mud take to drop the 460 m?
(b) How fast was it traveling when it reached the bottom?
(c) How long did it take the mud to slide the 8 km horizontally?
23. A flowerpot falls from the ledge of an apartment building. A person in the apartment below,
coincidentally holding a stopwatch, notices that it takes 0.200 sec for the pot to fall past his
window, which is 4 m high. How far above the top of the window is the ledge from which the
pot fell?
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24. Lou applies for a job as a perfume salesman. He tries to convince the boss to try his daring,
aggressive promotional gimmick: dousing perspective customers as they wait at bus stops. A
hard ball is to be thrown straight upward with an initial speed of 24.0 m/sec. A thin-skinned ball
filled with perfume is then thrown straight upward along the same path with a speed of 14 m/sec.
The balls are to collide when the perfume ball is at the high point of its trajectory, so that it
breaks open and everyone gets a free sample (whether they want it or not). If t = 0 when the
first ball is thrown, find the time when the perfume ball should be thrown.
25. Without telling Sally, Joe made travel arrangements that include a stopover in Toronto to visit
Joe’s old buddy. Sally doesn’t like Joe’s buddy and wants to change their tickets (for a small fee
of $100 per ticket). She hops on a courtesy motor scooter and begins accelerating at 0.9 m/sec2
toward the ticket counter to make the arrangements. As she begins moving, Joe is 40 m behind
her running at a constant speed of 9.0 m/sec.
(a) How long does it take for Joe to catch up with her?
(b) What is the time interval during which Sally remains ahead of Joe?
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26. A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/sec.
Air resistance may be ignored.
(a) At what time after being ejected is the boulder moving at 20.0 m/sec upward?
(b) At what time is it moving at 20.0 m/sec downward?
(c) When is the displacement of the boulder from its initial position zero?
(d) When is the velocity of the boulder zero?
(e) What are the magnitude and direction of the acceleration while the boulder is
i. moving upward?
ii. moving downward?
iii. at its highest point?
27. Suppose the acceleration due to gravity were only 0.98 m/sec2 instead of 9.8 m/sec2, but the
initial velocities with which you can jumpward upward or thrown a ball are unchanged.
(a) Calculate the height to which you could jump vertically from a standing start if you can jump
to 0.75 m when g = 9.8 m/sec2.
(b) How high could you throw a baseball if you can throw it 18 m up when g = 9.8 m/sec2.
(b) Calculate the maximum height of a window from which you would care to jump to a concrete
sidewalk below if for g = 9.8 m/sec2 you are willing to jump from a height of 2.0 m, which is the
typical height of a first story window.
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Projectile Motion:
Please add key points here from your Moodle Journal on Projectile Defined.
28. Another joyful physics student throws her cap in the air with an initial velocity of 24.5 m/sec at 36.9 degrees from the
horizontal. Find (a) the total time the cap is in the air and (b) the horizontal distance traveled.
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29. A policeman chases a master jewel thief across city rooftops. They are both running at 5 m/sec when they come to a
gap between the buildings that is 4 m wide and has a drop of 3 m. The thief, having studied a little physics, leaps at 5
m/sec and at 45 degrees and clears the gap easily. The policeman did not study physics and thinks he should maximize
his horizontal velocity, so he leaps at 5 m/sec horizontally. Does he clear the gap? By much how does the thief clear the
gap?
30. It’s the bottom of the ninth with two outs and the winning run is on base. You hit a knee-high fastball that just clears
the leaping third baseman’s glove. He is standing 28 m from you and his glove reaches 3.2 m above the ground. The flight
time to that point is 0.64 sec. Assume the ball’s initial height was 0.6 m. Find
(a) the initial speed and direction of the ball.
(b) the time at which the ball reaches its maximum height.
(c) the maximum height of the ball.
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31. A small steel ball is projected horizontally off the top landing of a long rectangular staircase. The initial speed of the
ball is 3 m/sec. Each step is 0.18 m high and 0.3 m wide. Which step does the ball strike first.
32. A baseball hit toward center field will land 72 m away unless it is caught first. At the moment the ball is hit, the center
fielder is 98 m away. He uses 0.5 sec to judge the flight of the ball, then races to catch it. The ball’s speed as it leaves the
bat is 35 m/sec. Can the center fielder catch the ball before it hits the ground?
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33. Wally and Luke advertise their circus act as “The Human Burrs - Trapeze Artists for the New Millenium.” Their
specialty involves wearing padded Velcro suits that cause them to stick together when they make contact in midair. While
working on their act, Wally is shot from a cannon with a speed of 20 m/sec at an angle of 30 degrees above the
horizontal. At the same moment, Luke drops from a platform having (x,y) coordinates of (8m, 16m), if the cannon is
taken to sit at the origin.
(a) Will they make contact?
34. A zoo-keeper, aims his tranquilizer gun directly at the belly of a monkey that is hanging in a tree. The zoo keeper fires his gun
from a height of 1.0 meter off the ground and the monkey’s belly is at a height of 11.0 meters off the ground. The zoo keeper is
positioned 20 meters away (horizontally) from the base of the tree. He fires his gun with an initial velocity of 25 m/s. At the same
instant that the gun is fired, the monkey releases his hands and falls from the tree, attempting to avoid being hit by the tranquilizer
gun. Will the monkey be hit? NOTE: Take all numbers to at least 4 sig figs!!!
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Uniform Circular Motion (taken from SparkNotes)
Uniform circular motion occurs when a body moves in a circular path with constant speed. For example, say you
swing a tethered ball overhead in a circle:
If we leave aside gravity for the moment, the only force acting on the ball is the force of tension, T, of the string. This force
is always directed radially inward along the string, toward your hand. In other words, the force acting on a tetherball
traveling in a circular path is always directed toward the center of that circle.
Note that although the direction of the ball’s velocity changes, the ball’s velocity is constant in magnitude and is always
tangent to the circle.
Centripetal Acceleration
From kinematics, we know that acceleration is the rate of change of the velocity vector with time. If we consider two points
very close together on the ball’s trajectory and calculate
radius of the circle.
, we find that the ball’s acceleration points inward along the
The acceleration of a body experiencing uniform circular motion is always directed toward the center of the circle, so we
call that acceleration centripetal acceleration,
. Centripetal comes from a Latin word meaning “center-seeking.”
We define the centripetal acceleration of a body moving in a circle as:
where v is the body’s velocity, and r is the radius of the circle. The body’s centripetal acceleration is constant in magnitude
but changes in direction. Note that even though the direction of the centripetal acceleration vector is changing, the vector
always points toward the center of the circle.
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How This Knowledge Will Be Tested
Most of us are accustomed to think of “change” as a change in magnitude, so it may be counterintuitive to think of the
acceleration vector as “changing” when its magnitude remains constant. You’ll frequently find questions on SAT II Physics
that will try to catch you sleeping on the nature of centripetal acceleration. These questions are generally qualitative, so if
you bear in mind that the acceleration vector is constant in magnitude, has a direction that always points toward the
center of the circle, and is always perpendicular to the velocity vector, you should have no problem at all.
Objects Released from Circular Motion
One concept that is tested frequently on SAT II Physics is the trajectory of a circling body when the force providing
centripetal acceleration suddenly vanishes. For example, imagine swinging a ball in a circle overhead and then letting it
go. As soon as you let go, there is no longer a centripetal force acting on the ball. Recall Newton’s First Law: when no net
force is acting on an object, it will move with a constant velocity. When you let go of the ball, it will travel in a straight line
with the velocity it had when you let go of it.
EXAMPLE
A student is standing on a merry-go-round that is rotating counterclockwise, as illustrated above. The
student is given a ball and told to release it in such a way that it knocks over the wicket at the top of the
diagram. At what point should the student release the ball?
When the student releases the ball, it will travel in a straight line, tangent to the circle. In order to hit the wicket, then, the
student should release the ball at point B.
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35. Determine the speed (in m/s) of a car which travels around a 47.7-m diameter circle in a time of 18.82
seconds.
36. A car moving at a speed of 65.3 mi/hr moves around a circle with a diameter of 39.2 meters. Determine the
acceleration (in m/s/s) of the car. (GIVEN: 2.24 mi/hr = 1.00 m/s)
37. Determine the acceleration (in m/s/s) of a rider on the Cajun Cliffhanger (a barrel ride at an amusement
park) if the rider makes 7.4 revolutions around the 6.53-m diameter circle in 26.1 seconds.
38. An eraser is tied to a string and swung in a circle with a radius of 0.887 meters. The eraser makes 104.9
revolutions in a minute. Determine its acceleration (in m/s/s).
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39. A tire 0.53 m in radius rotates at a constant rate of 226 revolutions per minute. Find the speed (in m/s) of a
small stone lodged in the tread of the tire (on its outer edge).
40. (Referring to the previous problem.) Determine the acceleration (in m/s/s) of the stone.
41. Young David who slew Goliath experimented with slings before tackling the giant. He found that with a
sling of length 0.63 m, he could develop a rotation rate of 8.34 rev/s in his weapon. If he increased the length to
0.926 m, he could revolve the sling only 6.28 times per second. What is the centripetal acceleration (in m/s/s) at
8.34 revolutions per second?
42. (Referring to the previous problem.) What is the centripetal acceleration (in m/s/s) at 6.28 revolutions per
second?
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