Chapter 2
Lecture Notes
The branch of physics that deals with objects in motion is called mechanics. Mechanics is
composed of kinematics (which describes how objects move) and dynamics (which describes
why objects move). In this chapter we study one-dimensional kinematics.
Motion can only be described with respect to some frame of reference. Picture a number line
where an object is at position x1 at time t1 and at position x2 at some later time t2. In SI units we
measure position in meters and time in seconds.
Define displacement x = x2 – x1 as the change in position of the object. Note this can be
positive or negative, representing motion in some direction. Quantities described by both a
number (magnitude) of units and a direction are best represented by mathematical objects called
vectors. We indicate vectors by bold type or arrows placed over the variable representing the
vector. The SI unit for displacement is the meter.
Displacement is not the same as distance. Distance is a scalar (described by magnitude only). It
is never negative. If an object moves from the origin to a position 3 m to the right and then
returns to the origin it has traveled a distance of 6 m but its displacement is 0 m. This is because
its initial position x1 = 0 m and its final position x2 = 0 m. For any round-trip, the displacement is
always zero.
x 2  x1
x
=
as the rate of change position. This is also a vector,
t
t 2  t1
positive or negative indicating direction. It is measured in SI units of m/s. It is not the same as
speed. Average speed is defined as distance divided by time and is not a vector. Speed is always
non-negative. In the previous example, if the object took 6 seconds to complete the round trip, its
average speed would be 6 m / 6 s or 1 m/s. Its average velocity would be 0 m / 6 s or 0 m/s.
Define average velocity v =
To describe the motion of an object more precisely we want to know more than just the average
velocity. Did the object always have this velocity, or did it go faster, then slower, pause or
change direction? By measuring the position of the object more frequently than just at the
beginning and end of its trip we can calculate its velocity over shorter intervals of time.
lim x
as the moment by moment velocity, measured
t  0 t
over infinitesimally short intervals of time. Calculus is the branch of mathematics dealing with
change. We will not use calculus in this course but calculus students will soon learn that
instantaneous velocity as defined here is the derivative of position. For the rest of us, just think
of this as average velocity over very short time intervals.
Define instantaneous velocity v =
The velocity of an object may change. It may go faster to the left (or to the right) or slower.
v  v1
v
Define average acceleration a = 2
=
as the rate of change of velocity. In SI units,
t
t 2  t1
m/s
average acceleration is measured in units of
or m/s2.
s
Chapter 2
Lecture Notes
Note if average acceleration is positive it does not necessarily mean an object is speeding up. If
an object is moving to the right (chosen here as the positive direction), a positive acceleration
means the velocity is becoming more positive so the object goes faster and faster as it moves to
the right. But if the object is moving to the left, with negative velocity, a positive acceleration
means the final velocity v2 is less negative than the initial velocity v1 so the object must be
moving more and more slowly to the left. In the same way, a negative acceleration does not
always mean an object is slowing down.
If acceleration and velocity have the same sign, the object is going faster. If acceleration and
velocity have opposite signs, the object is slowing down.
lim  v
in order to more accurately
t  0  t
measure the change in velocity of an object. Calculus students will learn acceleration thus
defined is the first derivative of velocity (with respect to time) and the second derivative of
position.
Finally, we can define instantaneous acceleration a =
Equations of Motion (one-dimensional, constant acceleration)
We now manipulate the previous definitions to find a set of equations useful in solving motion
problems. These equations apply only to one-dimensional motion with constant acceleration. If
acceleration is not constant, try to break the problem into parts where acceleration is constant.
Start each problem at time t = 0. The initial position (in m) of an object at this time, we write as
x0 to emphasize it is the original position at time 0. The velocity (in m/s) of the object at this time
we similarly write as v0. At some later time t, the object is at position x with velocity v.
The definition of average velocity gives v =
x  x0
, and the definition of average acceleration
t
v  v0
. Note we have written a instead of a since we are assuming acceleration is
t
constant. The acceleration at any instant is just the average acceleration. We also use the constant
v v
acceleration assumption to write v = 0
; the average velocity is just the plain old average
2
of the initial and final velocities when acceleration is constant. With some rewriting and
algebraic manipulation (carried out in full in our text) we arrive at the following set of equations.
gives a =
v = v0 + at
1 2
at
2
v2 = v02 + 2a(x – x0)
v v
v = 0
2
x = x0 + v t
x = x0 + v0t +
Chapter 2
Lecture Notes
Example
A Pontiac is traveling 30.0 m/s east when the driver hits the brakes. The brakes provide a
constant acceleration of 6.00 m/s2 west. Find the time and distance required to stop the car.
Take east to be positive. Let the initial position of the car x 0 = 0 m (or solve for x – x0). Then we
want to find t and x (actually we want distance but the final position will give us this). We are
given v0 = 30.0 and v = 0 (the car stops). The acceleration a = –6.00 m/s2 (negative because it is
west). Remember, when the velocity and acceleration have opposite signs the car will slow down
(sometimes this is called deceleration).
We can use our equations above since acceleration is constant and the motion is one-dimensional
v  v0
along the east-west direction. Since v = v0 + at, we solve for t and find t =
or
a
t = (0.00 m/s – 30.0 m/s) / -6.00 m/s2 = 5.00 s. Of course, we could have just realized a
deceleration of 6 m/s2 means the car was slowing at 6 m/s each second and would require 5 s to
stop from its initial 30.0 m/s. Note we have written our answer with 3 significant figures
corresponding to the given information. Don't forget to include units.
There are many ways to find x but a simple one is to note the average velocity of the car is 15.0
v v
m/s east and in 5.00 s it will travel 75.0 m where we have used v = 0
and x = x0 + v t.
2
Thus t = 5.00 s and x = 75.0 m. It never hurts to think about your answers to see if they are
reasonable. From 60 mph would it take about 260 ft to stop? All done.
Falling Bodies
Galileo, over 400 years ago, saw that in the absence of air resistance all bodies fall to earth with
the same (constant) acceleration. This acceleration is the acceleration due to gravity and near the
surface of the earth is g = 9.80 m/s2 = 32 ft/s2 (downward). The equations developed in this
chapter may be applied to this one-dimensional up and down motion just as we applied them to
the left-right motion of the car in the previous example. In the next chapter we consider motion
in more than one dimension.
We are free to choose up to be positive or down to be positive. If there is no upward motion, it is
usually easier to let down be positive. We may leave the equations in terms of x or replace x with
y if it helps in thinking of vertical motion.
Example
Sammy Sosa hits a baseball straight up with initial upward velocity 49.0 m/s and initial position
1.00 m above home plate. How high will the ball go and how long will it be before it returns to
its original position?
We will neglect air resistance and take up to be the positive direction. We write y0 = 1.00 m, v0 =
49.0 m/s, and a = -9.80 m/s2. Note the signs. Let y be the position of the ball at its highest point
Chapter 2
Lecture Notes
and t the time to rise to this position. The time to fall to its original height can be shown to equal
the time to rise so we want to find y and 2t (time to rise + time to fall being the time the ball is in
the air).
Note the velocity of the ball at its highest position is 0 m/s. (The acceleration is still -9.80 m/s2 at
this and all other positions). To find the time to rise use v = v0 + at or
0 m/s = 49.0 m/s + (-9.80 m/s2)t giving t = 5.00 s. We have substituted before solving here; in
more complicated problems it is better to solve algebraically before substituting. Notice we
included the units in our calculations.
The time to rise is 5.00 s so the ball takes 10.0 s before returning to the level of the bat. Its
velocity at that point will be 49.0 m/s downward if we neglect air resistance; these problems are
very symmetric.
To find the height attained by the ball we can use the average velocity formula as before or
1
y = y0 + v0t + at2. We have replaced the x by y in the equations of motion. Substitute (don't
2
forget a is negative) and find y = 123.5 or about 124 m to 3 significant figures. Is this
reasonable?
Graphical Analysis of Motion
Given a graph of position versus time (meaning position is the vertical axis and time the
horizontal axis) our definition of average velocity implies that average velocity over some time
interval is the slope of the line (or more generally, curve) between those times. Calculus students
will learn about slopes of curves but we will stick with slopes of lines. The slope 'at a point' is the
instantaneous velocity. Recall a line that rises to the right has a positive slope, a line that falls to
the right a negative slope, a flat line has zero slope, and a vertical line has undefined slope.
Given a graph of velocity versus time, the slope of this graph between any two times is the
average acceleration over that interval. The slope 'at a point' is the instantaneous acceleration.
The area between the curve and the x-axis between any two times is the displacement over that
interval. Areas above the x-axis are positive; below, negative.
Words of Advice
Memorize the equations of motion and be able to solve any problem involving horizontal or
vertical motion. Problem solving is a very important part of this course. See me at once if you
have any problems with this material.
Air resistance cannot always be neglected. Falling bodies do not continue to accelerate in the real
world but reach a terminal velocity that depends on the environment, size, and density of the
body. The human body reaches terminal velocity of about 125 miles per hour (200 kph) after
falling for 13 to 14 s over a distance of 1880 ft (573 m). Stewardess Vesna Vulovic is said to
have survived the destruction of a DC-9 flying over Yugoslavia on January 26, 1972. She fell
without a parachute for 10160 m or 33000 ft.
Chapter 2
Lecture Notes