- Mechanics KINEMATICS DYNAMICS - Electricity - Magnetism -
Optics - Waves 5 Main Branches of Physics A description of motion A
study of what causes motion
Slide 3
The purpose of this chapter is to learn the 1 st step of
Mechanics (the study of motion) which is KINEMATICS (the study of
motion with no regards to what is causing the motion). The study of
what is causing the motion is known as dynamics, and we will study
this in a later chapter.
Slide 4
Scalar vs. Vector Scalar a quantity that has a magnitude only,
no direction. Ex: time (5 hours) age(17 years) temperature (20C)
distance (20 miles) * YES, scalars have units. Vector a quantity
that has both magnitude AND a direction. Ex: displacement (10 m
[S]) force (5 N [E])
Slide 5
Distance vs. Displacement Distance (d) the length of the path
followed by an object (scalar) * If an objects path is straight,
the distance is the length of the straight line between start and
finish. ** If an objects path is NOT straight, the distance is the
length of the path if you were to straighten it out and measure it
the way you would measure the length of a curved shoelace. start
finish start finish
Slide 6
Using the number line above, what would be the distance
travelled if an object travelled from .. - A to B - A to C - A to C
and then back to A - C to B, passing through A BC -3 -2 -1 0 1 A
meters 1m 4m 4m + 4m = 8m 4m + 1m = 5m
Slide 7
Displacement the change in an objects position during a time
interval. (vector) OR the length of a straight line from start to
finish. Displacement distance. However, sometimes the magnitude
will be the same. It doesnt matter what path you take from your
house to school, displacement will never change but distance will.
*Displacement must have both a magnitude (size) and a direction
(right, left, up, down, north, south, etc).
Slide 8
Using the number line above, find the distance travelled and
the displacement in moving from - A to B - C to A - A to C and then
back to A - C to B, passing through A BC -3 -2 -1 0 1 A meters 1m,
1m [right]* *notice a direction is given for displacement 4m, 4m
[left]* 8m, 0m no direction needed here 4m, 3m [left]* x = 1 (1m) =
0m x = (-2) (1m) = -3m OR 3m [left] + or can be used instead of R
and L
Slide 9
Speed vs. Velocity Average Speed (s) the distance travelled
during a time interval divided by the elapsed time. (scalar) s =
dist/ t ( or s=dist/t)
Slide 10
Larry walks from point B to point C, and then goes directly to
point A. If he walks at an average speed of 6 mph, how long does
the trip take him? BC -3 -2 -1 0 1 A miles d = 3mi + 4mi = 7mi s =
6 mi/h s = d/t t = d/s = (7mi)/(6mi/h)=1.17h 1 h, 10 min Use
appropriate units
Slide 11
Larry runs from point A to point B in 5 minutes and then
proceeds to jog directly to point C, taking his time in 30
additional minutes. Find BC -3 -2 -1 0 1 A km a)Larrys average
speed during the first portion of the trip. b)The average speed
during the second portion of the trip. c)Larrys average speed for
the entire trip. s = d/t = (1km)/(5min) = 0.2 km/min = 12 km/h s =
d/t = (3km)/(30min) = 0.1 km/min = 6 km/h s = d/t = (4km)/(35min) =
0.114 km/min = 6.86 km/h
Slide 12
Average Velocity ( v ) the displacement of an object divided by
the elapsed time. (vector) Avg. velocity is a change in position
over a change in time. Since displacement distance, velocity speed.
v = displace/ t (or v=d/t) *This line means its a vector
Slide 13
AD B C Sam runs the 400m dash. He starts and finishes at point
A, travelling one complete circuit around the track. Each section
of the track is 100m long. His average speed during each interval
are as follows. AB: 7 m/s BC: 8 m/s CD: 6 m/s DA: 7.5 m/s s = d/t t
= d/s = 100m/7sec = 14.286 sec 100m/8sec = 12.5 sec 100m/6sec =
16.667 sec 100m/7.5sec = 13.333 sec s = d/t = (400m)/(56.786s) =
7.04 m/sec Find Sams avg. speed and avg. velocity for the entire
trip. Avg Velocity = 0 since x = 0 for the entire trip. He ended in
the exact location he started!!
Slide 14
AD B C Sam runs the 400m dash. He starts and finishes at point
A, travelling one complete circuit around the track. Each section
of the track is 100m long. His average speeds during each interval
are as follows. AB: 7 m/s, 14.286 sec BC: 8 m/s, 12.5 sec CD: 6
m/s, 16.667 sec DA: 7.5 m/s, 13.333 sec 31.831 m 100 104.94 Find
Sams average speed and average velocity for the 1 st half of the
race. s = d/t t = d/s = 200m/(14.286+12.5s) = 7.47 m/s v = x/t =
(104.94m )/(14.286+12.5s) = 3.92 m/sec Use Pythagorean theorem to
determine the displacement from A to C
Slide 15
Scalar vs. Vector What scalars have we learned about thus far?
distancespeedtime What vectors have we learned about thus far?
Displacementvelocity
Slide 16
Scalars vs. Vectors has magnitude & direction (example: 15
cm east) has a magnitude only (example: 6 ft) 1 2 A B Displacement
is NEVER greater than distance traveled! Displacement:
Distance:
Slide 17
Scalars vs. Vectors (continued) has magnitude & direction
(example: 15 mi/h North) has a magnitude only (example: 30 km/h) If
an object STARTS & STOPS at the same point, the velocity is
ZERO! (since the displacement is zero) Velocity: Speed: 1 2 24 km 7
km Total time for the trip from 1 to 2: 1 hr Speed = 31 km/h
Velocity = 25 km/h at 16 o NE 25 km 16 o Dont worry about this
notation for this test just give the general direction
Slide 18
Speed on a d-t graph can be found by taking the
_______________. This gives us the change in distance of an object
over a change in time. SLOPE s AB = rise/run = (30-0m) / (10-0s) =
3 m/s s CD = rise/run = (100-50m) / (20-15s) = 10 m/s t (sec) d (m)
B C E A D F 10 15 20 27 36 120 100 50 30 Graphing distance vs.
time
Slide 19
d-t graphs (distance-time) Constant speed Speeding UP notice
how more distance is covered each second Constant Speed (but faster
than AB) Slowing Downless dist. covered each second At rest no
distance covered, but time goes by t (sec) d (m) B C E A D F 10 15
20 27 36 120 100 50 30
Slide 20
520 170yd = 350 yd (approximately) FYI -d-t graphs CANNOT have
sharp points. That would mean you came to a stop instantaneously
without slowing down first. minutes Practice Graph
Slide 21
x-t graphs (position time graphs. Like d-t graphs) t (sec) x
(m) t 1 t 2 t 3 x2x1x3x2x1x3 B C D A Constant speed (Constant +
velocity, or constant velocity in the + direction) Slow down, speed
up, slow down, speed up 2 moments where the object is at rest (for
a moment) imagine slowing your car to a stop, then going in reverse
( displacement)
Slide 22
How to get the displacement/position (d) at a certain time (t)
off an d-t graph. Sometimes Ill refer to this as x- t d(m) 10 20 30
40 50 t (s) 30 20 10 0 Example: What is the position at t = 30
seconds? Go over to t = 30. Find the pt on the curve. Find the x
value for this time. 24m
Slide 23
How to calculate the displacement between two times on an x-t
graph x(m) 10 20 30 40 50 t (s) 30 20 10 0 Example: What is the
displacement from t = 10 to t = 40? Find x i Find x f Use x = x f -
x i = + 7 m 10 m 17 m
Slide 24
How to find the distance traveled between two times on an x-t
graph. x(m) 10 20 30 40 50 t (s) 30 20 10 0 Example: What is the
distance traveled from t = 10 to t = 40? Find the distance traveled
in the + direction to the highest point. Find the distance traveled
in the direction from the highest point. Add them together. (27 m)
17 m 10 m
Slide 25
Understand the difference between velocity and speed on an x-t
graph. x(m) 10 20 30 40 50 t (s) 30 20 10 0 Example: What is the
average speed from t = 10 to t = 40 seconds? dist 10-40 = 27 m
(previous slide) Avg. Speed = dist/ t = 27m / 30 sec = 0.9 m/s 17 m
10 m
Slide 26
Understand the difference between velocity and speed on an x-t
graph. x(m) 10 20 30 40 50 t (s) 30 20 10 0 Example: What is the
average velocity from t = 10 to t = 40 seconds? Avg. Velocity =
slope = x/ t = + 7 / 30 sec = + 0.23 m/s x 10-40 = + 7 m (previous
slide) Notice the + sign. It indicates direction.
Slide 27
Will avg. velocity EVER be greater than avg. speed? NO!!! Will
avg. velocity EVER be equal to avg. speed? YES!!! When the path
travelled was one-way, in a straight line.
Slide 28
Negative Average Velocity? x(m) 10 20 30 40 50 t (s) 30 20 10 0
Avg. vel. = slope = rise/run = -7 m / 20 = -.35 m/s Example: What
is the average velocity from t = 20 to t = 40 seconds? Since the
objects displacement is in the NEGATIVE direction, so is its
average velocity.
Slide 29
-10 m avg velocity = slope = -15m / 6sec = -2.5 m/s s = |v| =
2.5 m/s At rest at t = 0 and t = 12 sec Graph Example 2) 3) 4)
Definition Instantaneous Velocity (v) the velocity of an object
at a precise moment in time.
Slide 32
Just what is meant by instantaneous velocity? tt tt tt tt tt To
find the average velocity between two points in time, we find the
slope of the line connecting these two points, thus finding the
change in position (rise) over the change in time (run). As the two
points move closer together, we find the average velocity for a
smaller time interval. As the two points move EVEN CLOSER together,
we find the average velocity for an EVEN SMALLER time interval.
Finally, in the limit that the time interval is infinitely small
(or approximately zero), we find the velocity at a single moment in
time. Hence the term instantaneous velocity
Slide 33
To find instantaneous velocities, we still use the concept of
slope. But we find the slope OF THE TANGENT TO THE CURVE at the
time in question Definition Tangent to a Curve a line that
intersects a given curve at exactly one point.
Slide 34
Good Tangents Bad Tangents
Slide 35
How to find the instantaneous velocity of a specific time
interval from an x-t graph x(m) 10 20 30 40 50 t (s) 30 20 10 0
Draw the tangent to the curve at the point in question. Then, find
the slope of the tangent. Slope = rise/run = 15 m / 9 s = 1.7 m/s
(approx) Example: What is the instantaneous velocity at t = 20
seconds? YOU MUST SPECIFY WHICH POINTS YOU USED WHEN FINDING THE
SLOPE!!!! (24, 30) (15, 15)
Slide 36
How to find the instantaneous velocity of a specific time
interval from an x-t graph x(m) 10 20 30 40 50 t (s) 30 20 10 0
Example: What is the instantaneous velocity at t = 5? If the pt
lies on a segment, find the slope of the segment. Slope = 5 m / 10
s = 0.5 m/s (0,5) (10,10) YOU MUST SPECIFY WHICH POINTS YOU USED
WHEN FINDING THE SLOPE!!!!
Slide 37
How to find the instantaneous velocity of a specific time
interval from an x-t graph x(m) 10 20 30 40 50 t (s) 30 20 10 0
Draw the tangent to the curve at the point in question. Then, find
the slope of the tangent. Slope = 0 (object at rest) Example: What
is the instantaneous velocity at t = 25 seconds?
Slide 38
x-t graphs t (sec) x (m) t 1 t 2 t 3 x2x1x3x2x1x3 1 2 3 0 Slope
of line segment
Slide 39
Tangent to the curve has a slope of -26m / 13.5s = -1.93 m/s
THEREFORE, v = -1.93 m/s and s = 1.93 m/s (approximately) Open to
in your Unit 1 packet 1 Tangent to the curve has a slope of +22m /
22sec = 1 m/s (13.5,-20) (0,6) (33,2) (11,-20)
Slide 40
Definition Average Acceleration ( a ) the change in an objects
velocity in a given time interval..IN OTHER WORDS, the rate of
change of an objects velocity. When you stop a car, you actually
push the break petal a few seconds before coming to a complete
stop. The velocity gradually slows. a = v f -v iOR a = v/t t f -t
i
Slide 41
Find the acceleration of each object 1) An object is moving at
40 m/s when it slows down to 20 m/s over a 10 second interval. 2)
An object is moving at -40 m/s, and 5 seconds earlier it was moving
at -10 m/s. 3) An object travelling at -10 in/min is moving at +20
in/min 2 minutes later. 4) An object moving at -30 mph is moving at
-20 mph 10 hours later. a = v/ t = (20 40m/s) / 10sec = -2 m/s 2 a
= v/ t = [-40 (-10m/s)] / 5sec = -6 m/s 2 a = v/ t = [20 (-10
in/min)] / 2min = +15 in/min 2 Slows down Negative accel Speeds up
Negative accel Speeds up + Accel a = v/ t = [-20 (-30 mi/h)] / 10hr
= + 1mi/h 2 Slows down + Accel
Slide 42
How can something have a negative acceleration when traveling
in a positive direction? When a train, traveling in a positive
direction (right) slows as it approaches the next station, velocity
can still be +, but acceleration will be negative because initial
velocity is larger than final velocity. v is negative. But be
careful Negative acceleration doesnt always mean deceleration.
Think of a train moving in a negative direction (in reverse, or
just to the left). Acceleration would be negative when the train
gained speed and positive when the train lost speed to enter a
station.
Slide 43
Did the last line confuse you? How can something slow down and
have a positive acceleration? This example may help An object
moving at -30 mph is moving at -20 mph 10 hours later. Its speed
(30mph vs. 20 mph) clearly decreases. * remember, speed is |v| As
time marches on, the velocity become MORE positive (b/c -20mph is
more positive than - 30mph) THEREFORE, v is +
Slide 44
What do the units of acceleration mean? m/s 2 3 m/s 2 means
that your velocity increased by 3 m/s every second. -0.1 km/min 2
means that your velocity decreased by 0.1 km/min every minute that
you were moving. m/s/s m/s per second t (sec)v (m/s) 00 13 26 39
412 t (min)v (km/min) 00.3 10.2 20.1 30 4-0.1
Slide 45
The Key Equations Displacement: d = d f - d i Velocity:
Acceleration: AVERAGE INSTANTANEOUS Really just tangents to the
curve at a point.
Slide 46
-1 m/s1 m/s +4 m/s4 m/s +9 m/s9 m/s +14 m/s14 m/s +19 m/s19
m/s
Slide 47
Call right + and left v i = 5 m/s right = + 5 m/sv f = 4.8 m/s
left = -4.8 m/s a = (v f v i ) / t = (-4.8 5) /.002sec = -4,900 m/s
2 = 4,900 m/s 2 left v i = 60 mi/hv 2 = 0 mi/ha = -5 mi/h 2 a = (v
f v i ) / t -5 = (0 60) / t t = 12 s
Slide 48
x-t v-t END OF TODAYS LECTURE
Slide 49
x t UNIFORM Velocity Speed increases as slope increases x t
Object at REST x t Object Positively Accelerating x t Object
Negatively Accelerating x t Moving forward or backward x-t s x t
Changing Direction x t Object Speeding up
Slide 50
x t x t x t x t POSITIVE OR NEGATIVE ACCEL? SPEEDING UP OR
SLOWING DOWN? Slope of the tangent gives v inst Getting more sloped
speeding up Getting more + sloped + Accel Getting less sloped
slowing down Slopes are getting less + - Accel Getting less sloped
slowing down Slopes are getting less + Accel Getting more sloped
speeding up Slopes are getting more - Accel
Slide 51
An easy way to remember it Im Negative!!! Im Positive!!!
Slide 52
Find the acceleration in each case. v 1 = 10 m/s, v 2 = 20 m/s,
t = 5sec v 1 = 10 m/s, v 2 = -20 m/s, t = 10sec v 1 = -9 km/h, v 2
= -27 km/h, t = 3 h v 1 = -9 km/h, v 2 = 6 km/h, t = 3 h
Slide 53
v t UNIFORM Positive (+) Acceleration Acceleration increases as
slope increases v t UNIFORM Velocity (no acceleration) Object at
REST v t Changing Direction v-t s v t UNIFORM Negative (-)
Acceleration v t
Slide 54
v(m/s) 2 4 6 8 10 12 t (s) 8 6 4 2 0 -2 -4 v-t graphs Constant
+ accel (speeding up) Constant + Vel (constant speed) Constant
negative accel (slowing down) At rest Constant negative accel
(speeding up) Constant + accel (slowing down) Constant - Vel
Slide 55
v(m/s) 2 4 6 8 10 12 t (s) 8 6 4 2 0 -2 -4 How to get the
velocity (v) at a certain time off a v-t graph Example: What is the
velocity at t = 8 seconds? Go over to t = 8. Find the pt on the
graph. Find the v value for this time. -2 m/s
Slide 56
v(m/s) 2 4 6 8 10 12 8 6 4 2 0 -2 -4 Finding the average
acceleration on a v-t graph A 2-4 = (v f v i ) / t = rise / run = 0
m/s 2 A 4-10 = (v f v i ) / t = rise / run = -7 / 6 = -1.17 m/s 2
Example: What is the average acceleration between 0 & 2, 2
& 4, and 4 & 10 seconds? a 0-2 = (v f v i ) / t = rise /
run = +4/2 = +2 m/s 2
Slide 57
v-t graphs t (sec) v (m/s) Slope of any segment is the AVERAGE
acceleration The slope of the tangent to the curve at any point is
the INSTANTANEOUS acceleration t 0 t 1
Slide 58
30 20 10 0 -10 -20 -30 a = slope = (+57 m/s) / 32sec = +1.78
m/s 2 (approx) Open to in your Unit 1 packet 3 v = -30 m/ss = 30
m/s a = slope = (-30 m/s) / 16sec = -1.875 m/s 2
Slide 59
30 20 10 0 -10 -20 -30 Open to in your Unit 1 packet 3 You cant
say. You know its speed at the start, but not where it is Const
accel (object speeds up), const vel, const + accel (slows down),
const + accel (speeds up), const accel (slows down) END OF LECTURE
Object is at rest whenever it crosses the t-axis t = 0, 36, 80 sec
3) 4) 5)
Slide 60
x t UNIFORM Velocity Speed increases as slope increases x t
Object at REST x t Object Positively Accelerating x t Object
Negatively Accelerating x t Moving forward or backward x-t s x t
Changing Direction x t Object Speeding up
Slide 61
v t UNIFORM Positive (+) Acceleration Acceleration increases as
slope increases v t UNIFORM Velocity (no acceleration) Object at
REST v t Changing Direction v-t s v t UNIFORM Negative (-)
Acceleration v t
Slide 62
A Quick Review The slope between 2 points on an x-t graph gets
you the _______________. The slope at a single point (the slope of
the tangent to the curve) on an x-t graph gets you the
____________. The slope between 2 points on a v-t graph gets you
the ____________. The slope at a single point (the slope of the
tangent to the curve) on a v-t graph gets you the ____________.
Average velocity Inst. velocity Inst. accel. Avg. accel.
Slide 63
NEW CONCEPT When you find the area under the curve on a v-t
graph, this gets you the displacement during the given time
interval. This is NOT the area under the curve The area under the
curve is really the area between the graph and the t-axis. v t v
t
Slide 64
Find the area under the curve from . a) 0-4 seconds. b) 4-6 c)
6-10 d) 0-10 A = (4)(-10) = -20m v t 4 6 10 15 -10 A = (2)(-10) =
-10m A = (4)(15) = 30m A = -20 + (-10) + 30 = 0m The displacement
during the first 4 seconds is -20m The displacement during the next
2 seconds is -10m The displacement during the next 4 seconds is 30m
The OVERALL displacement from 0 to 10 seconds is zero (its back to
where it started)
Slide 65
v(m/s) 24 6 8 10 12 t (s) 8 6 4 2 0 -2 -4 How to find the
displacement from one time to another from a v-t graph Example:
What is the displacement from t = 2 to t = 10? Add the positive and
negative areas together x = 16 m + (-6.75 m) = 9.25 m Find the
positive area bounded by the curve 12 m + 4 m = 16 m Find the
negative area bounded by the curve (-2.25 m) + (-4.5 m) = - 6.75
m
Slide 66
v(m/s) 24 6 8 10 12 t (s) 8 6 4 2 0 -2 -4 How to find the
distance traveled from one time to another from a v-t graph
Example: What is the distance traveled from t = 2 to t = 10? Add
the MAGNITUDES of these two areas together distance = 16 m + 6.75 m
= 22.75 m Find the positive area bounded by the curve 12 m + 4 m =
16 m Find the negative area bounded by the curve (-2.25 m) + (-4.5
m) = - 6.75 m
Slide 67
v(m/s) 24 6 8 10 12 t (s) 8 6 4 2 0 -2 -4 How to find the
average velocity during a time interval on a v-t graph Example:
What is the average velocity from t = 2 to t = 10? The AVG velocity
= x / t = 9.25 m / 8 s = 1.22 m/s The DISPLACEMENT is simply the
area under the curve. x = 16 m + (-6.75 m) = 9.25 m 12 m + 4 m = 16
m (-2.25 m) + (-4.5 m) = - 6.75
Slide 68
v(m/s) 24 6 8 10 12 t (s) 8 6 4 2 0 -2 -4 How to find the final
position of an object using a v-t graph (and being given the
initial position) Example: What is the final position after t = 10
seconds if x i = 40 m? x = x f x i x f = x + x i = 13.25 m + 40 m =
53.25 m The DISPLACEMENT during the 1 st 10 sec is simply the area
under the curve. x = 20 m + (-6.75 m) = 13.25 m 4 m 12 m + 4 m = 20
m (-2.25 m) + (-4.5 m) = - 6.75
Slide 69
v-t graphs t (sec) v (m/s) Slope of any segment is the AVERAGE
acceleration The slope of the tangent to the curve at any point is
the INSTANTANEOUS acceleration t 0 t 1 The area under the curve
between any two times is the CHANGE in position (the displacement)
during that time period.
Slide 70
Open to in your Unit 1 packet 3 Displacement = |area| =
(16)(-30) + 8 (-30) = -480m v = x/t = -480 m / 24 sec = -20 m/s 6)
7) Distance travelled = |area| = | (16)(-30) | = 240m s = d/t = 240
m / 16 sec = 15 m/s 30 20 10 0 -10 -20 -30
Slide 71
Open to in your Unit 1 packet 3 Area = (16)(-30) + 12(-30) +
(8)(-30) + (8)(30) = - 600 m 8) Find all the areas under the curve
from 0 to 44 sec Area = x = - 600 m x = x f x i -600m = x f (-16m)
x f = - 616m 30 20 10 0 -10 -20 -30
Slide 72
Open to in your Unit 1 packet 4 +3.3 m/s 2 +10 m/s 0 m/s +75
m
Slide 73
-2 m/s 2 10 -10 0 m 50 m30 m 5
Slide 74
9) 10) 11) 12) 13) 14 & 34 sec + 2 m/s 2 +.35 m/s 2 + 8 m/s
approx 0.8 m/s 2