Kinematics in 1D and 2D
Book Chapter 2, and Sections 3-6 to 3-9
Motion in One
Dimension
Frames of Reference
All measurement must be made with respect to
a reference frame.
Example: if you are sitting on a train and someone walks
down the aisle, the person’s speed with respect to the
ground is much higher than with respect to the.
Distance vs Displacement
Physics Lingo
1) Displacement refers to how far an object is from its
starting point, regardless of how it got there.
Displacement is a vector.
2) Distance refers to the total length of the path
traveled by the object. Distance is a scalar.
y
x
ConcepTest 2.1
Walking the Dog
You and your dog go for a walk to the
park. On the way, your dog takes many
side trips to chase squirrels or examine
1) yes
fire hydrants. When you arrive at the
park, do you and your dog have the same
displacement?
2) no
ConcepTest 2.1
Walking the Dog
You and your dog go for a walk to the
park. On the way, your dog takes many
side trips to chase squirrels or examine
1) yes
fire hydrants. When you arrive at the
park, do you and your dog have the same
2) no
displacement?
Yes, you have the same displacement. Because you and your dog had
the same initial position and the same final position, then you have (by
definition) the same displacement.
Follow-up: have you and your dog traveled the same distance?
Velocity
Velocity is a measure of how fast the displacement
is changing.
Speed is a measure of how fast distance is being
traveled.
The instantaneous velocity is the average velocity
in the limit as the time interval becomes
infinitesimally short.
Instantaneous vs Average Velocity
The instantaneous
speed always equals
the magnitude of the
instantaneous velocity;
it only equals the
average velocity if the
velocity is constant.
Instantaneous vs Average Velocity (cont.)
On a graph of a particle’s position vs. time, the
instantaneous velocity is the tangent to the
curve at any point.
ConcepTest 2.3
Position and Speed
1) yes
If the position of a car is
2) no
zero, does its speed have to
3) it depends on
be zero?
the position
ConcepTest 2.3
Position and Speed
1) yes
If the position of a car is
2) no
zero, does its speed have to
3) it depends on
be zero?
the position
No, the speed does not depend on position; it depends on the change
of position. Because we know that the displacement does not depend
on the origin of the coordinate system, an object can easily start at x =
–3 and be moving by the time it gets to x = 0.
ConcepTest 2.5
Does the speedometer in a
car measure velocity or
speed?
Speedometer
1) velocity
2) speed
3) both
4) neither
ConcepTest 2.5
Does the speedometer in a
car measure velocity or
speed?
Speedometer
1) velocity
2) speed
3) both
4) neither
The speedometer clearly measures speed, not velocity. Velocity is a
vector (depends on direction), but the speedometer does not care
what direction you are traveling. It only measures the magnitude of
the velocity, which is the speed.
Follow-up: how would you measure velocity in your car?
Acceleration
Acceleration is the rate of change of the velocity.
Graphs of Kinematic Quantities
Equations of Motion
These are all the equations we need to solve
constant-acceleration problems in 1D.
ConcepTest 2.8b
Acceleration II
When throwing a ball straight up,
1) both v = 0 and a = 0
which of the following is true about
2) v  0, but a = 0
its velocity v and its acceleration a
3) v = 0, but a  0
at the highest point in its path?
4) both v 0 and a  0
5) not really sure
ConcepTest 2.8b
Acceleration II
When throwing a ball straight up,
1) both v = 0 and a = 0
which of the following is true about
2) v  0, but a = 0
its velocity v and its acceleration a
3) v = 0, but a  0
at the highest point in its path?
4) both v 0 and a  0
5) not really sure
At the top, clearly v = 0 because the ball has
momentarily stopped. But the velocity of the
ball is changing, so its acceleration is definitely
not zero! Otherwise it would remain at rest!!
Follow-up: …and the value of a is…?
y
Solving Problems
1. Read the whole problem and make sure you
understand it. Then read it again.
2. Decide on the objects under study and what the time
interval is.
3. Draw a diagram and choose coordinate axes.
4. Write down the known (given) quantities, and then
the unknown ones that you need to find.
5. What physics applies here? Plan an approach to a
solution.
Solving Problems (cont.)
6. Which equations relate the known and unknown
quantities? Are they valid in this situation? Solve
algebraically for the unknown quantities, and check
that your result is sensible (correct dimensions).
7. Calculate the solution and round it to the
appropriate number of significant figures.
8. Look at the result—is it reasonable? Does it agree
with a rough estimate?
9. Check the units again.
Freely Falling Objects
Near the surface of the Earth,
all objects experience
approximately the same
acceleration due to gravity.
The acceleration due to gravity
at the Earth’s surface is
approximately 9.80 m/s2.
Freely Falling Objects (cont.)
At a given location on the Earth and in the
absence of air resistance, all objects fall with the
same constant acceleration.
ConcepTest 2.9b
Alice and Bill are at the top of a
building. Alice throws her ball
downward. Bill simply drops
his ball. Which ball has the
greater acceleration just after
release?
Free Fall II
1) Alice’s ball
2) it depends on how hard
the ball was thrown
3) neither—they both have
the same acceleration
4) Bill’s ball
Alice
Bill
v0
vA
vB
ConcepTest 2.9b
Alice and Bill are at the top of a
building. Alice throws her ball
downward. Bill simply drops
his ball. Which ball has the
greater acceleration just after
release?
Free Fall II
1) Alice’s ball
2) it depends on how hard
the ball was thrown
3) neither—they both have
the same acceleration
4) Bill’s ball
Both balls are in free fall once they are
Alice
released, therefore they both feel the
v0
Bill
acceleration due to gravity (g). This
acceleration is independent of the initial
vA
vB
velocity of the ball.
Follow-up: which one has the greater velocity when they hit the
ground?
ConcepTest 2.12a
Throwing Rocks I
You drop a rock off a
1) the separation increases as they fall
bridge. When the rock
has fallen 4 m, you drop a 2) the separation stays constant at 4 m
3) the separation decreases as they fall
second rock. As the two
rocks continue to fall,
4) it is impossible to answer without more
what happens to their
information
separation?
ConcepTest 2.12a
Throwing Rocks I
You drop a rock off a
1) the separation increases as they fall
bridge. When the rock
has fallen 4 m, you drop a 2) the separation stays constant at 4 m
3) the separation decreases as they fall
second rock. As the two
rocks continue to fall,
4) it is impossible to answer without more
what happens to their
information
separation?
At any given time, the first rock always has a greater velocity than the
second rock, therefore it will always be increasing its lead as it falls.
Thus, the separation will increase.
ConcepTest 2.12b
Throwing Rocks II
You drop a rock off a
1) both increase at the same rate
bridge. When the rock
2) the velocity of the first rock increases
has fallen 4 m, you drop
faster than the velocity of the second
a second rock. As the
3) the velocity of the second rock
two rocks continue to
increases faster than the velocity of the
fall, what happens to
first
their velocities?
4) both velocities stay constant
ConcepTest 2.12b
Throwing Rocks II
You drop a rock off a
1) both increase at the same rate
bridge. When the rock
2) the velocity of the first rock increases
has fallen 4 m, you drop
faster than the velocity of the second
a second rock. As the
3) the velocity of the second rock
two rocks continue to
increases faster than the velocity of the
fall, what happens to
first
their velocities?
4) both velocities stay constant
Both rocks are in free fall, thus under the influence of gravity only.
That means they both experience the constant acceleration of gravity.
Since acceleration is defined as the change of velocity, both of their
velocities increase at the same rate.
Follow-up: what happens when air resistance is present?
ConcepTest 2.14b
v versus t graphs II
1) decreases
Consider the line labeled B in
2) increases
the v vs. t plot. How does the
3) stays constant
speed change with time for
4) increases, then decreases
line B?
5) decreases, then increases
v
A
t
B
ConcepTest 2.14b
v versus t graphs II
1) decreases
Consider the line labeled B in
2) increases
the v vs. t plot. How does the
3) stays constant
speed change with time for
4) increases, then decreases
line B?
5) decreases, then increases
v
A
t
B
In case B, the initial velocity is positive
but the magnitude of the velocity
decreases toward zero. After this, the
magnitude increases again, but
becomes negative, indicating that the
object has changed direction.
Example A
The position of a small object is given by
x = a + bt + ct3, where a, b, and c are constants.
Find the velocity and acceleration as a function of
time.
Using the definitions of velocity and acceleration:
dx d
3
v
 (a  bt  ct )
dt dt
v(t )  b  3ct
dv d
a
 (b  3ct 2 )
dt dt
a(t )  6ct
2
Example B
A sports car accelerates from rest to 100 m/s in
5.0 s. What is the average acceleration in SI units?
The definition of average acceleration is:
v final  vinitial
a
t
Applying this definition to our problem, we get:
(100  0) m/s
a
5.0 s
a  20 m/s 2
Motion in Two
Dimension
Vector Kinematics
In two or three dimensions,
the displacement is a
vector:
Vector Kinematics (cont.)
As Δt and Δr become
smaller and smaller, the
average velocity
approaches the
instantaneous velocity.
Vector Kinematics (cont.)
The instantaneous
acceleration is in the direction
of Δ v = v2 – v1, and is given
by:
Vector Kinematics (cont.)
Using unit vectors,
Summary of Equations
Generalizing the one-dimensional equations for
constant acceleration:
Projectiles in Physics
A projectile is an
object moving in two
dimensions under the
influence of Earth's
gravity; its path is a
parabola.
Projectile Motion
Projectile motion is motion in two dimensions,
with constant acceleration g downwards.
Parabolic Nature of Projectile Motion
Projectile motion is parabolic:
Taking the equations for x and y as a function of
time, and combining them to eliminate t, we
find y as a function of x:
This is the equation for a parabola.
Analyzing Projectile Motion
It can be understood by
analyzing the horizontal
and vertical motions
separately.
Analyzing Projectile Motion (cont.)
The speed in the x-direction
is constant; in the y-direction
the object moves with
constant acceleration g.
This photograph shows two balls that
start to fall at the same time. The one
on the right has an initial speed in the
x-direction. It can be seen that vertical
positions of the two balls are identical
at identical times, while the horizontal
position of the yellow ball increases
linearly.
Analyzing Projectile Motion (cont.)
If an object is launched at an initial angle of q0
with the horizontal, the analysis is similar except
that the initial velocity has a vertical component.
ConcepTest 3.4a
A small cart is rolling at
constant velocity on a flat
track. It fires a ball straight
up into the air as it moves.
After it is fired, what
happens to the ball?
Firing Balls I
1) it depends on how fast the cart is
moving
2) it falls behind the cart
3) it falls in front of the cart
4) it falls right back into the cart
5) it remains at rest
ConcepTest 3.4a
A small cart is rolling at
constant velocity on a flat
track. It fires a ball straight
up into the air as it moves.
After it is fired, what
happens to the ball?
In the frame of reference of
the cart, the ball only has a
vertical component of
velocity. So it goes up and
comes back down. To a
ground observer, both the
cart and the ball have the
same horizontal velocity,
so the ball still returns into
the cart.
Firing Balls I
1) it depends on how fast the cart is
moving
2) it falls behind the cart
3) it falls in front of the cart
4) it falls right back into the cart
5) it remains at rest
when
viewed from
train
when
viewed from
ground
ConcepTest 3.4b
Now the cart is being pulled
along a horizontal track by an
external force (a weight
hanging over the table edge)
and accelerating. It fires a ball
straight out of the cannon as it
moves. After it is fired, what
happens to the ball?
Firing Balls II
1) it depends upon how much the
track is tilted
2) it falls behind the cart
3) it falls in front of the cart
4) it falls right back into the cart
5) it remains at rest
ConcepTest 3.4b
Now the cart is being pulled
along a horizontal track by an
external force (a weight
hanging over the table edge)
and accelerating. It fires a ball
straight out of the cannon as it
moves. After it is fired, what
happens to the ball?
Firing Balls II
1) it depends upon how much the
track is tilted
2) it falls behind the cart
3) it falls in front of the cart
4) it falls right back into the cart
5) it remains at rest
Now the acceleration of the cart is completely unrelated to the ball. In
fact, the ball does not have any horizontal acceleration at all (just like
the first question), so it will lag behind the accelerating cart once it is
shot out of the cannon.
ConcepTest 3.7b
Punts II
A battleship simultaneously fires two shells at two enemy
submarines. The shells are launched with the same initial
velocity. If the shells follow the trajectories shown, which
submarine gets hit first ?
1
3) both at the same time
2
ConcepTest 3.7b
Punts II
A battleship simultaneously fires two shells at two enemy
submarines. The shells are launched with the same initial
velocity. If the shells follow the trajectories shown, which
submarine gets hit first ?
The flight time is fixed by the
motion in the y-direction. The
higher an object goes, the longer
it stays in flight. The shell hitting
submarine #2 goes less high,
therefore it stays in flight for less
time than the other shell. Thus,
submarine #2 is hit first.
1
3) both at the same time
2
Example 3-16 – Relative Velocity of Cars at 90°
Two automobiles approach a street corner at right
angles to each other with the same speed of 40.0
km/h (= 11.1 m/s), as shown. What is the relative
velocity of one car with respect to the other? That
is, determine the velocity of car 1 as seen by car 2.
Problem 3.32
A ball is thrown horizontally from the roof of a
building h meters tall, and lands R meters from
the base. What was the ball’s initial speed?
First, let us draw a detailed diagram of the situation.
y
v0
x
h
g
R
Problem 3.32 (cont.)
y
v0
x
h
g
R
Notice that the initial velocity has
no vertical component. Also, we
have chosen the ball’s initial
position to be at the origin of our
coordinate system, this will help
simplify things later.
Since this is projectile motion, we can use the projectile
motion equations for this problem:
x  x0  vx0t
y  y0  vy 0t  12 gt 2
From our diagram, we see that x0 = 0, and y0 = 0. Also,
we know that vy0 = 0. Now our equations simplify to:
Problem 3.32 (cont.)
y   12 gt 2
x  vx0t
Now we need to find vx0, since v0 = vx0 î. We can do this
by solving for t using the equation for x, to get:
x
t
vx 0
Next, we plug that result into the equation for y, and
rearrange it to get:
gx2
y 2
2v x 0
2y
vx 0   2
gx
Finally, we plug in the given values:
2(h)
vx 0  

2
gR
2h
2
gR
v0 
2h ˆ
i
2
gR
Problem 3.37
You buy a plastic dart gun, and you decide to find its
maximum horizontal range. You shoot the gun straight
up, and it takes 4.0 s for the dart to land back at the
barrel. What is the gun’s maximum horizontal range?
To find the gun’s maximum horizontal range, we first need to know
how fast the gun can shoot the darts.
We can use the information about the vertical shot to find the initial
speed of the dart.
v0 y
g
Since the dart is in free fall after it leaves
the gun, then its trajectory must be given
by:
y (t )  y0  v0t  12 gt 2
Problem 3.37 (cont.)
In this case, we are told that the dart starts from the barrel and
comes back to the same point in 4.0 s, so if we choose this point as
our origin, it makes y = y0 = 0. Then our equation simplifies to:
v0t0  12 gt02  0
Dividing both sides by t0, and solving for v0, we get:
v0  12 gt0
Were t0 = 4.0 s. Now that we know how fast the gun can shoot a
dart, we can use that velocity to calculate the maximum horizontal
range.
y
g
v0
q
x
R
The maximum range in projectile
motion occurs when the initial
velocity is q = 45º above the
horizontal.
Problem 3.37 (cont.)
Now we have enough information to write down the equations of
motion for the x and y components:
x(t )  (v0 cos q )t
y (t )  (v0 sin q )t  12 gt 2
Since we are looking for x = R, we can use the equation for y(t) to
find how long it takes for the dart to fall to the ground. Setting
y(t) = 0, we get:
0  (v0 sin q )t  12 gt 2
This is a quadratic equation with two solutions: t = 0 (which is the
time when the dart was first shot), and
t
2v0 sin q
g
We finally have the time it takes for the dart to land. Now all we
have to do is substitute this value into x(t), simplify, and that will
give us the value of R.
Problem 3.37 (cont.)
Thereby,
2v0 cos q  v0 sin q
R
g
2v02 cos q sin q

g
Knowing that sin 45  cos 45 
v0 obtained earlier, we get:
R
2( 12 gt0 ) 2
1
2
, and using the value of
1
2

1
2
g
gt02
R
 39 meters
4