ACCELERATED MOTION
Let’s begin by summarizing some of our earlier
definitions.
If position is represented by a vector x, then
displacement is change in position (in m, cm, km etc,
with direction), i.e.
x  x f  x i .
(Average or instantaneous) velocity is the displacement
divided by the elapsed interval of time (in m/s, cm/s,
km/h etc, with direction), i.e.
x x f  x i
v

t
t f  ti .
(Average or instantaneous) acceleration is the change in
velocity divided by the elapsed interval of time (in m/s2,
cm/s2, km/hr2, etc, with direction), i.e.
v v f  vi
a

t t f  t i .
For simplicity, let’s suppose we can assume that
acceleration is uniform, or constant. Then, since the
acceleration doesn’t change, “average acceleration” and
“instantaneous acceleration” are identical and we can
just refer to “acceleration”. Rearranging the last
equation will give
at  v  v f  vi , or (adding vi to both sides)
v f  vi  at .
[1]
2
Equation [1] shows us what the final velocity will be if we
know the initial velocity, the acceleration and the elapsed
time, and it can also be used to find one of the other
quantities if we know the remaining three. (We’ll see
examples of this shortly.)
Now, if the acceleration is constant the average velocity
can actually be written in two ways:
vi  v f
x
v
v
and
.
t
2
It follows that
 v  vf
x  vt   i
 2

 v  vi  at 
 2v  at 
t   i
t   i
t ,
2
2





or
x  vi t 
1
2
at  .
2
[2]
Essentially, equations [1] and [2] are all you need to solve
any problem about bodies moving with constant
acceleration!
Example 1
A car starts from rest and accelerates uniformly to the
east at 2 m/s2. Find its velocity and its displacement
after 8 seconds.
Solution
Since it starts from rest, vi = 0 m/s. So, from [1] and [2]
v f  vi  at  0  28  16 m/s (east) and
3
x  vi t 
1
1
2
2
at   08  28  64 m (east).
2
2
Example 2
A ball starts from rest and accelerates down a 6m ramp,
reaching the bottom in 4 seconds. What is its velocity at
the bottom?
Solution
First, using [2] to find the acceleration,
1 2
1
2
x  vi t  a t  , so 6  04   a 4  8a ,
2
2
3
i.e. a  4 m/s2 (down the ramp). Then, from [1], the
velocity at the bottom of the ramp is
3
v f  vi  at  0  4   3 m/s (down the ramp).
4
Example 3
The maximum braking acceleration of a car is –3 m/s2
(opposite to the direction of motion). What is the
braking displacement from 18 m/s (north) to rest and
from 36 m/s to rest?
Solution
First, the car starts at vi = 18 m/s (north) and ends at
vf = 0 m/s. So, using [1], we can find how long it takes to
stop, i.e.
v f  vi  at so 0  18   3t and t  6 s .
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From [2], it follows that the displacement during braking
is
x  vi t 
1
1
2
2
at   186   36  108  54  54
2
2
meters (north).
Repeating when vi = 36 m/s (north),
v f  vi  at so 0  36   3t and t  12 s .
Therefore,
x  vi t 
1
1
2
2
at   3612   312  432  216  216
2
2
meters (north).
Note that doubling your starting speed results in a
quadrupling of your stopping distance!
FREE FALL NEAR THE EARTH’S SURFACE
It turns out that the acceleration of gravity near the
Earth’s surface is virtually constant. Specifically,
measurements of gravitational acceleration taken at
thousands of locations around the globe would produce
results which vary only from about –9.78 m/s2 (down) to
–9.83 m/s2 (down). For this reason, we take the average
of these and say that the magnitude of the acceleration of
gravity here on Earth is g = 9.81 m/s2. (In fact, to keep
calculations simple, we often just use g = 9.8 m/s2.)
This means that equations [1] and [2] above can also be
used to describe objects in “free fall” near the Earth’s
surface, as long as we replace a by –g = –9.8 m/s2 (9.8
m/s2 down).
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Example 4
A stone is dropped from rest at a height of 60.0 m. How
long does it take to hit the ground and what is its velocity
at impact?
Solution
Since the stone falls 60.0 m, its displacement is –60.0 m.
Using equation [2],
1
2
1
x  vi t  a t  , so  60.0  0t   9.8t 2  4.9t 2 .
2
2
2


t
Hence

 60.0
 12.24 and t  12.24  3.5 s.
 4.9
Now, from [1], the stone’s velocity when it hits the
ground is
v f  vi  at  0   9.83.5  34.3 m/s (34.3 m/s
down) .
Any problem dealing with free fall or uniformly
accelerated motion can be solved using equations [1] and
[2], so make sure you have these equations on your
information sheet for the next test.
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NEWTON’S LAWS
Aristotle had believed (mistakenly) that a force was
required not only to start something moving but also to
keep it moving. Much later, Galileo realized that, in
fact, all bodies have a tendency to keep doing what they
are doing, i.e. if at rest, they will tend to remain at rest,
and if moving, they will tend to keep moving. This
tendency is today referred to as inertia, and it is the basis
for Newton’s First Law of Motion. (In the law as stated
below, the term net external force simply means the net
effect of all forces acting from outside a body.)
Newton’s First Law
Every body remains at rest or moves with constant velocity
except when compelled to change its motion as a result of
a net external force acting upon it. (Note that “constant
velocity” means both constant speed and constant
direction, i.e. the motion is steady and in a straight line.)
Think of a book at rest on a desk. It is easy to identify
two external forces which act vertically on the book:
gravity pulling down on it and the desk pushing up on it.
But, because the book remains at rest, we can conclude
there is no net external force, i.e. these two external
forces are equal and opposite, and hence cancel one
another. This serves to illustrate that a body can remain
at rest when no net external force is acting.
Now suppose the desk was frictionless and infinitely
long! Even a slight push (say to the right) for just a brief
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moment would get the book moving, but, with no friction
to slow it down and stop it, it would continue moving to
the right forever (i.e. long after the slight push had
ended), even though the same two external vertical
forces are still acting. This illustrates that a body can
continue to move uniformly with no net outside force on
it.
If we return to the book at rest on the desk, and
suddenly take the desk away, there is no longer an
upward push to oppose the downward pull of gravity.
That is, there will now be a net external (downward)
force on the book due to gravity, which explains why the
book will neither remain at rest nor move uniformly.
Instead, it will accelerate downward.
Finally, imagine that you are riding in a car travelling
east at 30 km/hr when the car suddenly hits a wall and
stops. Since you are not part of the car, your body’s own
inertia will make it tend to keep moving forward even
after the initial impact. That is, your body will keep
travelling east at 30 km/hr until an external force acts to
stop it. If you’re lucky, that external force will be
exerted by your seatbelt and perhaps your airbag. If
you’re unlucky, it will be exerted by the dashboard and
the windshield!
Momentum
Momentum is one of two additional physical quantities
associated with motion (the other being kinetic energy,
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which will be discussed later). Specifically, there are
actually two kinds of momentum: linear momentum
(which is associated with linear motion) and angular
momentum (which is associated with rotational or
spinning motion). In this discussion, we consider only
linear momentum, which is denoted p and defined to be
the product of mass and velocity. That is,
p  mv (with units kg-m/s) .
(Note that, because velocity is a vector and has a
direction, then linear momentum is also a vector, and
has the same direction as the velocity.)
Basically, this formula tells us that both a very small
mass with a very large velocity and a very large mass
with a very small velocity can possess a lot of
momentum. Ask yourself if you’d want to stand against
a wall in front of either an approaching slow moving
transport truck or an approaching speeding bullet and
you’ll get the idea! (Refer to Table 5.2 in the textbook
for some examples of magnitudes of linear momentum.)
Force
We have already introduced the idea of force as
something which feels like a “push” or a “pull”. The
term force is referred to specifically in Newton’s First
Law as the “agent” responsible for causing a body to
change its state of rest or motion with constant velocity.
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Since it is obvious that the effect of a push or pull can
depend on its direction, it is clear that force is a vector.
We sometimes divide everyday forces into two
categories: “contact” forces which result from bodies (or
particles within bodies) “touching” one another; and
“action-at-a-distance” forces, which can occur between
bodies even when are physically separated.
Suppose we return to a book sitting on a desk. The
downward gravitational pull of the Earth on the book is
clearly “action-at-a-distance” while the supporting force
of the desk (upward) against the book is considered a
“contact” force. (Other so-called “contact” forces
include tension in a string and friction which can either
keep a body from starting to slide across a supporting
surface or slow an already-sliding body until it finally
stops.)
In fact, however, all forces are explained in terms of the
four fundamental forces identified earlier—gravity, the
electromagnetic force, the weak nuclear force and the
strong nuclear force, all of which are really “action-at-adistance”!
As we’ve mentioned before, for example, the upward
push of a desk on a book is actually due to the
electrostatic repulsion between the electrons in the atoms
on the bottom of the book and the electrons in the atoms
on the top of the desk. And, because these electrons
never actually touch each other, then, when considered
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microscopically, the book and the desk are not really in
“contact” at all! In that sense, “contact” forces are a
myth.
If a net external force can cause a body to change its
state of rest or motion with constant velocity, it stands to
reason that such a force will change the body’s
momentum. This idea inspired Newton’s Second Law of
Motion.
Newton’s Second Law (First Version – universally true)
The net external force acting on a body is equal to the rate
at which its linear momentum changes, with the direction
of the momentum change being in the direction of the
force.
Quantitatively, this law can be written
p mv
F

t
t ,
where we have to remember that F is not interpreted as
a single force on any body, but rather as the net external
force, i.e. the net effect of all outside forces on a body.
The unit associated with force can be seen from the
equation to be the unit of momentum divided by time.
This is kg-m/s divided by s, or kg-m/s2, but because force
is so important in physical science, this combination has
been named the Newton (N), so that
1 N = 1 kg-m/s2 .
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In most (though not all!) situations, when a net external
force acts on a body to change its momentum, the mass
stays constant while the velocity changes. In such cases,
we can replace the momentum change (mv) by mv (to
reflect the fact that only the velocity actually changes).
That is,
mv mv
F

 ma ,
t
t
(since acceleration has already been defined by
v v f  vi
a

t t f  t i .)
This results in a second, and certainly better known
(though limited), version of Newton’s Second Law.
Newton’s Second Law (Second Version – true provided
mass is constant)
When a net external force acts upon a body of constant
mass, the body will experience an acceleration which is
proportional to the force (and in the same direction) and
inversely proportional to the mass, i.e.
F
a
F  ma [4] .
m [3] or
Before we consider examples involving this more
familiar version of Newton’s Second Law, one situation
where this version would not apply is the case of a
rocket. As you have certainly witnessed if you’ve
watched a space shuttle launch, rockets propel
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themselves forward by expelling large amounts of fuel
out the back. This means that the mass of a rocket in
fact decreases (usually very rapidly), and certainly can’t
be assumed to be constant. Thus, in applying Newton’s
Second Law to a rocket, one would have to use the First
p mv
F


Version,
t
t . (Unfortunately, the First
Version is considerably more difficult to use,
mathematically, and we won’t be able to discuss it
further in this course.)
Example 1
A net external force of 50 Newtons (right) is applied to a
2 kg mass at rest. Determine its velocity after 4 seconds.
Solution
Using equation [3],
a
F 50

 25 m/s2 (right) .
m 2
We can now find the velocity after 4 seconds from
equation [1],
i.e. v f  vi  at  0  254  100 m/s (right) .
Example 2
How much force is required to stop a 5,000 kg truck if it
is brought to rest from 30 m/s (east) in 10 s?
Solution
From [1], if v f  vi  at , then 0  30  a10 .
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So,
 30
a

 3 m/s2 (3 m/s2 west) .
10a  30 , or
10
It follows from [4] that the required force is
F  ma  5000 3  15000 N (15,000 N west) .