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Acceleration
Review: Velocity is the rate at which position changes.
Acceleration is the rate at which velocity changes. If your velocity is changing at all,
then you’re accelerating. Remember, your velocity can change if your speed changes or
if your direction changes. For example, you could be moving at a constant speed, but if
you turn a corner then your direction changes. So, by definition your velocity would
change, too. When that happens, you’re accelerating. Your acceleration is simply how
fast your velocity is changing.
In math form, a =
v
t
where a = the average acceleration
v = vf – vi
t = the time interval
Acceleration is a vector. Therefore, it must have a numerical value and a direction.
Until we get to Unit 3 (2-D Motion) we’ll just use +/- to describe the direction of the
acceleration.
If the acceleration is positive, the object is either:
a) getting faster as it’s moving forward in the positive direction.
Ex: if an object is going forward in the positive direction and its initial
velocity (vi) was +10 m/s and it went to +20 m/s (vf) in 5 seconds (t), then its
acceleration will be +2 m/s2.
20m / s  10m / s
5s
= +2 m/s2
Notice the units of acceleration are a “length unit” (m) over a “time unit
squared” (s2) or meters per second per second. More on this later…
*
b) getting slower as it is moving forward in the negative direction.
Ex: if an object is moving forward in the negative direction and its velocity
went from
-20m/s to –10m/s in 2 seconds, then its acceleration will be +5m/s. (Plug it
in the equation and do the math)
If the acceleration is negative, it means the car is either
a)
b)
slowing down while it’s moving in the positive direction
OR speeding up while it’s traveling in the negative direction. To verify
this, just pick any two numbers and use them as a vi and vf and pick some
time interval and then do the math.
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So, you need to think about the car’s motion when interpreting the sign of the
acceleration.
FYI: When a problem says “a car is moving and then it stops” its vf is 0 m/s. If a
problem says that a car “starts from rest”, its vi = 0 m/s.
Objects in this class will undergo something called “constant acceleration”. Any objects
undergoing “constant acceleration” are having their velocity change in a consistent way
(by a constant amount) during each time interval. For example, if a rocket has a constant
acceleration of +500m/s2, its velocity is changing by +500 m/s every single second of
time (that’s what the “meters per second per second” units mean). So, if its velocity was
initially 0m/s, 1s later its velocity would be +500m/s. One second after that its velocity
would be +1000m/s. One second later, its velocity would be +1500 m/s, etc.
Or, if an object has a constant acceleration of –10m/s/s, its velocity will get more and
more negative as time goes on. For example, let’s say an object was moving with an
initial velocity of +30m/s and with an acceleration of –10m/s/s (meters per second per
second, AKA meters per second-squared). One second later, its velocity will be +20m/s.
One second after that, it will +10m/s. One second later it will be 0m/s. Get it?
Pretty much every problem in this class will assume the object in any acceleration
problem is undergoing “constant acceleration”. When the problem says the object is
undergoing constant acceleration (they all will, trust me), there are four additional
equations that we’ll use to describe the object’s motion. These equations are just
derivations of the two equations you already know. Here they are:
1) vf = vi + (a)(t)
2) d =
1
( vf + vi)t
2
3) d = (vi)(t) +
1 (a)(t2)
2
4) (vf)2 = (vi)2 + 2(a)(d)
Your book might use slightly different equations and variables. Sometimes they might
call “d” (if it’s a vertical “d”) “y”, and sometimes they might call “d” (if it’s a horizontal
“d”) “x”. Some books give displacement the symbol “s”. I still can’t figure the logic
behind that one. We’ll always use just “d” for displacement. Know the four equations
as I’ve written them here for you…that’s the form I’ll be teaching them in.
vi = initial velocity; the velocity an object has at the beginning of some motion.
vf = final velocity; the velocity an object has at the end of some motion.
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d = total displacement (change in position; d; “d2-d1”) during the time interval “t” ; also
represents the total displacement the object had when it went from vi to vf ; Remember:
This is NOT always the same as the total distance traveled;
a = the acceleration of the object; units are m/s/s or m/s2; this is the acceleration that
makes the object’s velocity change from the Vi to the Vf.
t = the time interval the displacement takes place; also represents the time it took the
object to accelerate from its initial velocity to its final velocity; this is the same as t is the
first two equations you learned
Here are some practice problems:
A car traveling at +44 m/s is uniformly accelerated to a velocity of +22m/s over an 11 sec interval. What is
its displacement over this time?
Vi = +44m/s
vf = +22m/s
one other thing
t = 11s
d=?
The only equation that you can use is :d = ½ (vf + vI)t
It’s the only one that has everything you have PLUS the
you’re looking for.
A car accelerates at a constant rate from 15m/s to 25m/s while it travels 125m. How long does this motion
take?
vi =15m/s
vf = 25 m/s
d = 125 m
t=?
d = ½ (vf + vi) t
A bike rider accelerates at a constant rate to a velocity of 7.5m/s during 4.5s. The bike’s displacement is
+19m. What was the initial velocity of the bike?
vf = 7.5 m/s
t = 4.5 s
d = +19m
vi = ?
d = ½(vf + vi) t
An airplane starts from rest and accelerates at a constant rate of +3m/s/s for 30s before leaving the ground.
What is its displacement during this time?
vi = 0m/s (rest)
AND your
a = +3m/s2
other
t = 30s
that
The equation that has everything you have
unknown is d = vit + 1/2at2. If you look at the
equations, they all have some other variable in them
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d=?
you don’t have, so the above equation is the only
option.
A car is initially sliding backwards down a hill at –25km/h. The driver guns the car. By the time the car’s
velocity is +35km/h, it is +3.2m from its starting point. Find the car’s acceleration (assume it’s constant).
vi = -25km/h
WATCH YOUR UNITS!!!! The only option here
vf = +35km/h
d = +3.2 m
a=?
vf2 = vi2 + 2ad.
is:
A spaceship traveling at a velocity of +1210m/s is uniformly accelerated at –150m/s/s.
If the acceleration lasts for 8.68s, what is the final velocity of the craft? Really think
about what’s going on in this problem.
Vi = +1210 m/s
a = -150 m/s2
t = 8.68 s
vf = ?
Again, to figure out which equation to use, you must know the values for everything in
the equation except the one thing you’re looking for (your unknown). It must have your
unknown in it!!!!!
ACCELERATION DUE TO GRAVITY
When objects are freely-moving in the air (no jet engines making them constantly go up
against the force of gravity), they are under a single force (called gravity) that is
constantly pushing down on them. The force will cause the velocity of freely falling
objects to constantly change. This results in a form of constant acceleration called
gravitational acceleration. We refer to objects that are freely falling as objects in “freefall”. For example, if you throw a ball up in the air, it has the same acceleration acting on
it on the way up as it does on the way down. This acceleration will cause it to slow down
as it moves up in the air, and speed up as it is in the falling stage.
Gravity is a force that only points down (I’ll explain this further in Unit 2). So, the
acceleration that this “force of gravity” produces also points down (toward the Earth). In
this class, anything pointing down is assigned a negative value. So, the value given to the
“acceleration due to gravity” is always, always, always negative. Its value is: - 9.8 m/s2.
Objects in free-fall will never have any other acceleration as long as they’re at sea-level
on Earth (acceleration due to gravity changes with your distance away from the center of
the planet).
Things on the ground can control their acceleration (for example, a car can control its
acceleration by stepping on the gas and going faster or stepping on the brake to slow
down). So, things moving on the ground will have a bunch of different values of
acceleration. However, things in the air/atmosphere in free-fall (a ball, etc) will
ALWAYS have an acceleration of –9.8 m/s2 (even it it’s initially moving up in the air the
acceleration will still have this negative value…).
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When doing problems with objects flying through the air in free-fall, you’ll use the same
four equations that you just learned (the funky ones with a lot of variables). However,
whenever you have the variable “a”, you simply substitute the number “-9.8 m/s2”.
Sometimes you might see those equations with a “g” substituted for the “a”. The symbol
“g” just stands for acceleration due to gravity (but it means the same thing as “a”).
Everything else in the equations represent the same things that they did in the problems
you’ve already done.
In your problems, if Vi or Vf are negative, it simply means the object is moving DOWN.
Also, if the object is moving DOWN, the “d” (d) is negative as well. In this class, we’ll
always stick with the convention that anything moving DOWN is given a negative
value and anything moving UP is given a positive value. Some things might start off
going UP in the air (like a ball that is thrown up) and the velocity would begin as a
positive value. However, anything in free-fall will eventually come back down. When
that happens, for a split instant at the TOP of its flight its velocity will be zero. Then it
will begin to come down. When it begins its motion downward, its velocity will then
have a NEGATIVE value.
More problems for you to do on your own for practice….
If you drop a ball, how far does it fall in ½ sec?
Vi = 0m/s (it starts from REST—you’re just dropping it)
t = 0.5s
a (a=g) = -9.8m/s2
d =?
The only equation that has everything that you know AND
your unknown is
d = v i t + ½ a t2
A man falls 1m to the floor. How long does the fall take? How fast is he going when he
hits the floor?
If a man just “falls” it means that he has an original velocity of zero and his velocity
begins to change at a rate of –9.8 m/s2. So, his velocity will gradually become more
negative and negative as time passes. Since he is falling, his displacement is “down” and
d would be a negative number (in this case –1m).
d = -1m
solve for “t”,
Vi = 0m/s
huh????
a = -9.8m/s2
t =?
Vf = ?
For this problem, you will first use one equation to
then you will use another equation to solve for Vf. Fun,
On a wet pavement, a car can be accelerated with a maximum acceleration a = 0.20g before its tires slip. If
it started from rest, how fast is it moving after 2.0 s? After 4.0s?
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a = 0.2(9.8m/s2)
as a “standard”
Vi = 0m/s
acceleration
t = 2s
of 9.8 m/s/s.
Vf = ?
directional
d=?
Sometimes they will use acceleration due to gravity
to compare other accelerations. In this problem, a car’s
is 0.20 “g”. This means the car’s acceleration is only 20%
There is no need for the negative sign here as that is just a
tool. Since the car is clearly speeding up in the problem
(with a positive velocity), a positive acceleration is used.
A pitcher throws a baseball straight up with an initial speed of 27m/s. How long does it
take the ball to reach its highest point? How high does the ball get at its highest point?
Vi = +27 m/s (positive because it’s going UP)
a = -9.8 m/s2
t=?
*at its highest point, its velocity is zero. We can label that highest point as Vf. If
we do that, you will be looking for the time between when it left the ground with
its initial velocity and reached its highest point where its velocity was zero.
Vf = 0m/s
*For the second question, you’re asked to calculate the “d” between its lowest
point (with a velocity of Vi) and its highest point (where its velocity was Vf =
0m/s).
d=?
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