PES 1110 Fall 2013, Spendier
Lecture 3/Page 1
Review
Last time we talked about the first two physical quantities of kinematics:
Position, x - Where an object is located, how far and what direction from origin
- displacement: ∆x = x2 – x1
- distance: d = |x2|+ |x1|
Velocity, v - How fast an object is going and direction of motion. Slope of the positionversus-time graph.
displacment ∆x x2 (t2 ) − x1 (t1 )
m
Average velocity: vavg =
=
=
units:  
elapsed time ∆t
t2 − t1
s
Instantaneous velocity: time derivative of position v = lim
∆t → 0
∆x dx
=
∆t
dt
Today: continue chapter 2
- Acceleration
- Equations of motion (1 d motion constant acceleration)
Mathematical means of describing the relationships among these variables
Acceleration
Acceleration: a is the rate at which the velocity changes
What does a physicist mean when they say the word rate? It means divide by time.
Velocity is the rate at which position changes: so it is position divided by time.
a avg =
∆v v 2 -v1
=
∆t t 2 -t1
m/ s  m 
units : 
= 2
 s   s 
m/s per every second – most descriptive way - how much did the velocity change by
Example:
A world’s land speed record was set by Colonel John P. Stapp when on March 19th 1954,
he rode a rocket-propelled sled that moved down a track at 1020 km/h. He and his sled
were brought to a stop in 1.4s. Assuming constant acceleration, what acceleration did he
experience?
t1 = 0 s, t2 = 1.4 s
v1 = 1020 km/h, v2 = 0 m/s
need to convert km/h into m/s
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PES 1110 Fall 2013, Spendier
v 2 =1020
Lecture 3/Page 2
km 1000 m   1hr 
×
×
 = 283.3m / s
hr  1km   3600 s 
then
aavg ,r =
∆v v 2 -v1 0m / s − 283.3m / s
=
=
= −202.4m / s 2
∆t t 2 -t1
1.4 s − 0 s
He experienced a negative acceleration. Negative acceleration indicates acceleration in
the opposite direction to motion. This usually indicates breaking or deceleration.
We can also express this result in terms of g = 9.8 .8 m/s2 (magnitude of the average
acceleration due to earth’s gravity) The force of gravity that pulls you toward the earth. In
fact, acceleration forces are measured in g-forces (more in chapter 5), where 1 g is equal
to the force of acceleration due to gravity near the Earth's surface (9.8 m/s2, or 32 ft/s2).
What is his acceleration in g?
 1g 
−202.4m / s 2 
= −20.7 g
2 
 9.8m / s 
The deceleration was 20.7 g
He did not die - however a constant 15 g-s for a minute can be deadly.
Some g-forces we experience:
Rollercoaster: 2.7 g
a sneeze - 2.9 g
car staring 0.5 g, (Formula 1 car 1.7 g)
car cornering 0.9g (Formula 1 car 1-3g)
Example: A rabbit accelerates from rest to 2.2m/s in 0.5 s. Find the rabbit’s average
acceleration
t1 = 0 s, t2 = 0.2 s
v1 = 0 m/s, v2 = 2.2 m/s
aavg ,r =
∆v v 2 -v1 2.2m / s − 0m / s
=
=
= 4.4m / s 2
∆t t 2 -t1
0.5s − 0 s
Instantaneous acceleration
So to make this instantaneous at one instant of time, we have to find the slope at this
time, the slope of what graph?
Position vs time: slope gives velocity
Velocity vs time: slope gives acceleration
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PES 1110 Fall 2013, Spendier
Lecture 3/Page 3
Again we are doing this magnifying business
The instantaneous velocity (or simply acceleration)
∆v dv
a= lim
=
∆t → 0 ∆t
dt
When you read the book (section 2.6) you can see that we can combine what we know
about instantaneous velocity
dv d
d  dx  d 2 x
a= = v =   = 2
dt dt
dt  dt  dt
The acceleration of a particle at any instant is the second derivative of its poison x(t) with
respect to time.
%%%%%%%%%%%%% aside
What makes a rollercoaster thrilling....is it the fast speed? How fast does it go - 70 mph (i
looked it up) - well - I can go 70 miles per hour in my car and it is not fun
So what makes a rollercoaster ride exciting? It is acceleration. More precisely it is the
change in acceleration. The change in acceleration is caller “jerk” and defined as:
da d
d  dv  d  d  dx  d 3 x
Jerk =
= ( a ) =   =    = 3
dt dt
dt  dt  dt  dt  dt  dt
Jerk is the third derivative of position with respect to time. When rollercoaster are
designed engineers look at jerk. What instrument is used to measure jerk – a jerk meter.
%%%%%%%%%%%%%% end aside
Direction (and therefore) sign of acceleration
There is one more thing about acceleration…a tricky thing… what about the plus and
minus for acceleration. Velocity is easy – what does a positive velocity mean? Going to
the right. What does a negative velocity mean? Going to the left. Make my position
positive, where should I stand? To the right.
The direction (and therefore) sign of acceleration is more complicated than position’s or
velocity’s. You cannot say that a positive acceleration means speeding up and a negative
means slowing down.
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PES 1110 Fall 2013, Spendier
Lecture 3/Page 4
From our equation:
∆v
aavg ,r =
∆t
we see that acceleration is in the same directions as the change in velocity.
Example:
Let’s say we have this moving object. The object moves to the right and slows down:
From 10 m/s to 5m/s in a time ∆t:
v1 = 10 m/s
v2 = 5 m/s
Then the average velocity
∆v v 2 -v1 5m / s − 10 m / s
5m / s
aavg ,r =
=
=
=−
∆t
∆t
∆t
∆t
This is an example where the object moves into the positive direction but the acceleration
is negative, since the object is slowing down.
In General:
When a and v have the same sign, speed increases (mean absolute value increases).
When a and v have the opposite sign, speed decreases (mean absolute value
decreases).
Keep all these in mind as we get into problem solving
Constant Acceleration
Acceleration is constant with time.
Example: Vertical motion:
Near the surface of the earth, gravity can be considered as providing a constant 1D
acceleration acting towards the center of the earth. We will talk about gravity a lot more
later in the course but mention it here while we are looking a 1D acceleration.
Real life: What do you notice when you throw up a ball. If we throw a ball straight up, it
turns around at some point and comes back done due to gravity – motion graphs?
Position vs time graph corresponds to a parabola.
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PES 1110 Fall 2013, Spendier
Lecture 3/Page 5
Demonstration:
- Drop a whiteboard eraser and flat piece of paper simultaneously.
observation: paper falls more slowly
- Drop a whiteboard eraser and a crunched up piece of paper simultaneously.
observation: both objects fall at the same rate
Inference: After removing effects of air resistance, all objects accelerate downwards at
the same rate.
Gravitational acceleration: given by symbol g
On earth: g = - 9.8 m/s2
By definition, up is positive and since objects accelerator down, g is negative
All calculations in this course will ignore air resistance! So for free fall problems:
constant acceleration
A simplification in notation:
We can always assume that the initial time t1 starts at zero
Then elapsed time: t2-t1 = is t2 = t hence I do not need a subscript
∆t = t2 − t1 = t2 − 0 = t
Now by making this choice along comes a change in notation
call the final position x2 = x
call the initial position x1 = x0
The initials – will have a subscript of zero – to indicate to others and ourselves that we
made this choice that the initial time is zero.
Better notation:
v2 = v
x2 = x
v1 = v0
x1 = x0
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PES 1110 Fall 2013, Spendier
Lecture 3/Page 6
For example, with this new notation, the equation for average velocity becomes
x − x x − x0 x − x0
v avg = 2 1 =
=
t2 − t1
t −0
t
Equations of motion (along straight line for constant acceleration)
Mathematical means of describing the relationships among position, velocity, and
acceleration.
When the acceleration is constant, the average acceleration and instantaneous
acceleration (see velocity vs time graph) are equal and we can write
v - v0 v - v0
v −v
a = a avg = 2 1 =
=
t2 − t1
t −0
t
We can now solve for v
v(t) = v 0 +at
Eq. 1
our first Eq. of motion ==> missing x-x0
Test for t = 0, v = v0 as it should be
So in this first equation, we have no information about x or x0. How can we relate Eq. 1
to position? Through average velocity:
∆x x2 (t ) − x1 (t ) x(t ) − x0 (t )
=
=
∆t
t2 − t1
t −0
v avg =
Eq.2
For a linear velocity function - due to constant acceleration- the average velocity over any
time interval is the average at the beginning - v0 - and the velocity at the end - v -then for
interval t = 0 to t the average velocity is
v avg =
1
( v0 + v)
2
Eq.3
sub Eq.3 into Eq.2 and using Eq.1
x(t) - x 0 (t) 1
1
1
= ( v 0 + v ) = ( v 0 + [ v 0 + at ]) = v 0 + at
t
2
2
2
Hence:
1
x(t ) − x0 (t ) = v0 t + at 2
2
Eq. 4
2nd Eq. of motion ==> missing v
our parabola! Test: for t = 0, x = x0 as it should be.
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PES 1110 Fall 2013, Spendier
Lecture 3/Page 7
Then we can use some more algebra by combining Eq.4 and Eq. 1 to obtain other
equations of motion like:
v 2 = v0 2 + 2a ( x − x0 )
1
x(t ) − x0 (t ) = vt − at 2
2
1
x(t ) − x0 (t ) = ( v0 + v ) t
2
Eq. 5
3rd Eq. of motion ==> missing time
Eq. 6
4th Eq. of motion ==> missing v0
Eq. 7
5th Eq. of motion ==> missing a
Table 2-1 in the book has all these summarized. We have five possible equations (see
boxed ones), each involving three of the motion variables. They are valid in cases of
constant acceleration only.
Book section 2-8
In case you have already taken calculus - the equations can be derived in a different way.
As a physicist it is always useful to be able to derive something in more than one way.
Now we need to practice using these equations:
Example:
A fighter plane must reach a speed of 75 m/s from rest in a distance of just 75m. What
acceleration is required to achieve this?
picture:
knowns: v0 = 0m/s, v = 75 m/s, x0 = 0m, x = 75 m
unknowns: time and acceleration
to find a, we can use:
v 2 = v0 2 + 2a ( x − x0 )
( 75m / s ) - ( 0m / s )
v 2 - v 20
a=
=
= 37.5m / s 2 = 3.8 ×101 m / s 2
2 ( x-x 0 )
2 ( 75m - 0 )
2
2
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PES 1110 Fall 2013, Spendier
Lecture 3/Page 8
Example:
A car is traveling on a straight road with a speed of 30.0m/s when the driver hits the
brakes causing a constant deceleration of 2.5m/s2. How long does it take and how far
does the car go while stopping?
Step 1: draw a picture
Step 2: List of known variables
The initial position: x0 = 0 (always determined by your picture)
Initial velocity: v0 = 30 m/s
Acceleration: a = - 2.5 m/s^2 (arrows of velocity vectors for same time interval get
smaller)
final velocity: v = 0
Step 3: List of unknowns
Time t – how long does it take
Final position x – how far
Step 4: Use appropriate equations of motion
a) How long does it take the car to stop?
v = v0 + at
I know everything else but time inside of it
v − v0
0m / s − 30m / s
=t =
= 12 s
a
−2.5m / s 2
It takes the car 12s to stop.
b) How far does the car go?
1
1
x = x0 + ( v0 )t + at 2 = 0 + (−30 m / s )(12 s ) + (−2.5m / s 2 )(12 s )2
2
2
1
= 360 m + (−2.5m / s 2 )144 s 2 = 360 m −180 m = 180 m
2
So the car travels 180 m until it stops.
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PES 1110 Fall 2013, Spendier
Lecture 3/Page 9
Example:
A ball is thrown vertically upwards at a speed of 25.0 m/s.
a) At what time does the ball reach its maximum height?
t = 2.55 s
b) What is the ball's maximum height?
x = 31.9 m
c) How long is the ball in the air for if it is caught at the same height from which it was
thrown?
t = 5.1 s
d) What is the ball's velocity as it is caught?
v = 25 m/s down
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