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Unit 3: Physics – Motion (24 hours) |
The qualitative relationships among displacement, velocity and acceleration will be studied, and used to solve simple problems. Everyday phenomena and technologies will be discussed in terms of the motions involved.
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Ontario
Curriculum objectives: |
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Text: Science 10, Nelson
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Lesson One |
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Summary: (D7) |
Homework: |
PHYSICS
Physics: the study of matter and energy
-this
science deals with everything from smallest quarks and atoms to largest
galaxies
Quantitative
measurements of distances and times will be used often in this unit of the
course.
Qualitative Measurements are subjective
e.g. that baby is cute, it will take a while to go to the store
Quantitative Measurements in physics involve a number and units
e.g. that baby is 3 kg, it will take 17 minutes to go to the store
S.I.
=System International is the standardized metric measurement with units based
on 10’s
S.I.
base units for volume = L (litres)
mass = kg (kilograms)
time = s (seconds)
distance = m (metres)
S.I. (metric) Prefixes:
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k (kilo) |
h (hecta) |
da (deca) |
Standard Units |
d (deci) |
c (centi) |
m (milli) |
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1000 |
100 |
10 |
L, g, s, m |
0.1 |
0.01 |
0.001 |
A prefixes
in front of a standard unit alters the value of the standard unit:
e.g. 1 kL = 1000 L, 1 dg = 0.1 g, 1 das = 10 s, 1 mm = 0.001 m
Conveting
Units (review):
-in order to convert from one unit to another, start at the first prefix and
move the decimal to the second prefix in correct direction on the above chart.
e.g. 0.045 hg = ? dg
The prefix d (deci) is three prefixes to the right
of the prefix h (hecta), so move the decimal place three spaces to the right.
0.045 hg = 45 dg
e.g
9876.88 mm = ? m
The standard unit m (with no prefix) is three
prefixes to the left of the prefix milli, so move the decimal place three
spaces to the left.
9876.88 mm = 9.87688 m
Other
examples:
0.0837 kL = 8370 cL (move decimal five spaces to the right)
73 000 ds = 730 das (move decimal two spaces to the left)
0.1 dam = 0.001 km (move decimal two spaces to the left)
SCIENTIFIC NOTATION
Scientific Notation
is a simple way of writing large numbers
Write
the first number/decimal/other significant numbers/ x 10 exponent
eg. 6
230 000 000 000 000 000 = 6.23 x 1018 (first number is 6, decimal,
23 are significant, x 1018 as decimal was moved spaces 18 to left)
eg. 0.000 000 1 = 1 x 10-7 (first
number is 1, decimal, no other significant numbers, x 10-7 as
decimal was moved 7 spaces to right)
Other examples:
459.88 = 4.5988 x 102 (move decimal 2 spaces to the left)
0.0062707 = 6.2707 x 10-3 (move decimal 3 spaces to the right)
12 = 1.2 x 10 (move decimal 1 space to the left)
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Lesson Two |
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Summary: (D6,D7) |
Homework: |
SIGNIFICANT
FIGURES
Significant
Figures are those numbers that are considered valid numbers in an equation
-all
whole numbers and zero’s are significant except:
zeros to right of last whole number (e.g. 900 only the 9 is considered significant
and would be written as 9 x 102)
exception: zeros to right of decimals are considered significant (e.g.
8.00 has three significant figures)
Other
examples:
0.001 has 1
significant figure and is written as 1 x 10-3
1.00 has 3 significant figures
20 has
1 significant figures and is written as 2 x 10
202 has 3 significant figures and is written as 2.02 x 102
Do significant figure worksheet, questions 1 and 2 as a class.
When adding
or subtracting Significant Figures the answer should have least number
of decimal places
eg. 2.5
+ 1.35 + 2.382 = 6.232, but is written as 6.2 (2.5 only has one decimal, so the
answer should only have one decimal)
For
rounding from 5, round to nearest EVEN number
e.g. 10.5 rounds off to 10 if no decimals are needed. 9.5 also rounds off to 10
if no decimals are needed.
When multiplying
or dividing Significant Figures the answer should have least
number of significant figures
eg.
12.1 X 23.70 = 286.77, but is written as 287 (12.1 only has 3 significant
figures, so the answer should have only have three significant figures)
Do significant figure worksheet, questions 3 and 4 as a class.
KINEMATICS
Kinematics
is the study of motion
-there
are 2 main types of motion
1: Uniform Motion
-movement
in a straight line at constant speed eg. light, train, car
2: Non-Uniform Motion (accelerated)
-movement involving a change in direction and/or speed
eg. orbit, roller coaster
SCALARS and VECTORS
Two types
of measurements are Scalars and Vectors
1: Scalars
-a
quantity that has only magnitude (size) but no direction
eg. 90 km, 50 g
2: Vectors
-a
quantity that has both magnitude (size) and direction
eg. 90
km South
Some examples of scalars and vectors in kinematics: -a bunny hops 40 m East and 30 m South
Distance hopped is 40 m + 30 m = 70 m. This measurement does not indicate a
direction. DISTANCE is a SCALAR measurement.
Displacement
hopped is 50 m [east 37o]. This measurement does indicate a
direction. DISPLACEMENT is a VECTOR measurement.
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Lesson Three |
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Summary: (U1) |
Homework: |
SLOPE
Slope =
Rise / Run
-find
line of best fit on graph
-choose
any two points on a line (y1 , x1) (y2 , x2)
Slope =
Rise / Run
= (y2 – y1) / (x2
– x1)
= /\ y / /\ x where /\ means change
For a
distance vs. time graph, put distance on y axis, time on x axis
The slope of a line on this graph will be equivalent to the average speed of
the object measured
Slope = (d2
– d1) / (t2 – t1)
= /\d / /\t = vav (symbol for
scalar measurement called average SPEED)
Instantaneous
Speed = Speed calculated for one particular moment in time
Constant
Speed = Speed that does not change from point to point (this is needed for
motion to be considered Uniform Motion)
Do lab examinining d and vav
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Lesson Four |
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Summary: (U1,U2) |
Homework: |
DISPLACEMENT
-vector
measurements have a magnitude/size (with units) and direction
eg. 73 m [N] is a vector measurement with a magnitude of 73 m and a direction
of North
-distance is a scalar quantity (/\d) with no direction
change in
distance (/\d) = distance 2 (d2) – distance 1 (d1) is the
formula used to find a change in distance
/\d = d2 – d1
-displacement is a vector measurement indicating a change in
position, where position is written = d
(vectors symbols are written in bold or have --\ over the symbol
to distinguish them from scalar symbols)
displacement
(/\d) = position 2 (d2) – position 1 (d1)
is the formula used to find a change in position
/\d = d2
– d1
eg. I
I
I
House Library
Store
0 m 20 m
40 m
W<---------->E
For the
vector problems directions must be indicated.
If only two opposite directions are used make one positive and the other
negative when doing the math and convert back at the end.
For this example we will make East + and West - (This should be indicated
before the work is started)
change in
distance from the house to store: /\d = d2 – d1
=
40 m – 0 m
=
40 m, indicating that the store is 40 m from the house
displacement
from the house to store: /\d = d2
– d1
=
40 m – 0 m
=
40 m, this positive number indicates an easterly direction
=
40 m [E], indicating that the store was 40 m east of the house
change in
distance from the store to library: /\d = d2 – d1
=
20 m – 0 m, as your starting distance is always 0
=
20 m, indicating that the library is 20 m from the store
displacement
from the store to library: /\d = d2
– d1
=
20 m – 40 m, as these positions are fixed
=
-20 m, this negative number indicates a westerly direction
=
20 m [W], indicating that the library was 20 m west of the store
Resultant change in distance from house to store to library:
/\dR = /\d2 + /\d1 (add each change in individual distance to
find the total)
= 40 m + 20 m
= 60 m is the total distance travelled from the house to
the library
Resultant
change in displacement from house to store to library: /\dR
= /\d2 + /\d1 (add each individual displacement to find the total)
= (-20 m) + 40 m
= 20 m, this positive number
indicates an easterly direction
= 20 m [E] is the total
displacement from the house to the library
When
drawing vectors:
1. Place direction marker on page to indicate NSEW (north, south, east, west)
2. Draw a line to stated a scale, or write size next to vector (e.g. 1 cm drawn
= 1 m actual)
3. Use
an arrow to indicate the direction of the vector (each vector can only go one
direction)
For angles, write as 2m [70o E of N] which would
indicate the vector was 2 m long and moves at a straight line 70o to
the east of a northern direction.
This same vector could be written as 2m [20o N of E]
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Lesson Five |
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Summary: (U1,U2,D10) |
Homework: |
ADDING VECTORS
Resultant
displacement is found using the following formula /\dR = /\d1
+ /\d2 + /\d3 ... (sum of displacements)
Adding
Vectors to find resultant displacement can be done three ways:
1.
Using Scale Diagrams
2. Using Algebra
-state
directions
NSEW
-state which direction is + or –
-state
Given and
Required
-state Given and Required
-state
scale
-write equation for adding vectors
-draw
initial vectors to scale,
with -substitute
numbers with signs, solve
vectors
tail joining head of
last
-convert direction from + or -
-draw /\dR
from tail of first vector
to
head of last, and measure.
-convert
from scale units to actual units
eg.
Stephen walks 300 m [W] and then back 215 m [E] before stopping to talk to
a neighbor. Calculate the resultant
displacement.
North
Scale
diagram method: Scale: 1 cm = 50
m
W ^ E
/\d1
= 300 m [W]
/\d2
= 215 m [E]
/\dR
= ?
/\d1
<----------------
---------><-----
/\d2
/\dR
/\dR = 1.7 cm (50 m/1
cm)
= 85 m [W]
Algebra method: W =
negative, E = positive
/\d1
= 300 m [W] = -300 m
/\d2
= 215 m [E] = 215 m
/\dR
= ?
/\dR= /\d1
+ /\d2
= (-300 m) + (215 m)
= -85 m
= 85 m [W]
The two methods can also be combined, including all work of Algebra
method, and a relative (not scale) diagram.
Do pg. 424,
and analysis questions (a)-(g) to show vectors do not change direction,
and /\dR = /\d1 + /\d2 + /\d3
regardless
of order of vectors.
/\dR = /\d1 + /\d2 + /\d3
= /\d2 + /\d1 + /\d3
= /\d3 + /\d1 + /\d2
= /\d1 + /\d3 + /\d2
= /\d2 + /\d3 + /\d1
= /\d3 + /\d2 + /\d1
ADDING 2-DIMENSIONAL VECTORS
Do using
same format as scale diagram method, except include measurement of
angle for resultant displacement.
eg. A car goes 30 km [E] and 10 km [N]. Find the resultant
displacement. North
Scale: 1 cm = 10
km
^
/\d1
= 30 m [E]
/\d2 = 10 m [N]
/\dR
= ?
/\dR
l /\d2
/\d1
/\dR = 3.16 cm (but
remember that 1 cm in the drawing is equal to 10 km in reality)
= 31.6 km [17o N of E] or 31.6 km [73o E of N] (the
angle is found using a protractor)
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Lesson Six |
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Summary: (D8) |
Homework: |
VELOCITY
speed =
distance/time (a scalar measurement) velocity =
displacement/time (a vector measurement as it has a direction)
v =
/\d//\t
v = /\d//\t
eg. A train
goes 150 km [E] in 2 hours. What is the velocity of the train?
/\d = 150 km [E]
/\t = 2 h
v = ?
v = /\d//\t
= 150 km [E] / 2 h
= 75 km/h [E]
Average
Velocity = change in position from start to finish over a certain period of
time
vav = /\dR//\t (resultant
displacement/time)
eg. Monarch
butterflies are displaced 3500 km [SW] in 91 days.
What is the average velocity in km/h?
/\dR= 3500 km [SW],
/\t = 91 days (24 h/ 1 day) = 2184 h
vav= ?
vav= /\dR//\t
= 3500 km [E] / 2184 h
= 1.6 km/h [E]
AVERAGE SPEED VS. AVERAGE VELOCITY
A man walks
52 m east in 10 seconds, and then returns 41 m west in 8 seconds.
Vector
diagram:
North
52
m
\
^
\/
.
/\dR
41 m
Determine
the speed moved in each direction:
Speed
moved east:
v = /\d//\t
= 52 m / 10 s
= 5.2 m/s
Speed moved west:
v = /\d//\t
= 41 m / 8 m
= 5.1 m/s
Average speed:
vav = /\dR//\t
=(/\d2 + /\d1)/(/\t2 + /\t1)
= (52 m + 41 m)/(8 s + 10 s)
= 93 m/18 s
= 5.2 m/s
Find the
resultant displacement, and average velocity.
Make
East = negative, West = positive
/\d1
= 52 m [E]
= -52 m
/\d2 = 41 m [W]
= 41 m
/\dR
= /\d1+ /\d2
= -52 m + 41 m
= -11 m = 11 m [E]
vav = /\dR//\t
=(/\d1+ /\d2)/(/\t2
+ /\t1)
= (-52 m + 41 m)/(8 s + 10 s)
= 11m [E]/ 18 s
= 0.61 m/s [E]
Average Speed vav is NOT ALWAYS equal to Average Velocity vav
Describe graphs on page 447 of text.
Start car
racing lab with parts of class, while rest of class does pg. 436, #6,7
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Lesson Seven |
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Summary: (D2) |
Homework: |
ACCELERATION
acceleration = a = /\v//\t = rate of change of speed
eg. If a
car goes from a stop to 9.0 m/s in 3.0 s, what is the acceleration?
v2 = 9.0 m/s
v1 = 0.0 m/s
/\t = 3.0 s
a = ?
a = /\v//\t
= (v2 –v1)//\t
= (9.0 m/s - 0.0 m/s)/ 3.0 s
= 3.0 m/s2
aav = /\v//\t = (v2 –v1//\t
eg. If a
car is going 1.8 m/s, and 4.2 s later is going 8.3 m/s, what is the
acceleration of the car?
v2 = 8.3 m/s
v1 =1.8 m/s
/\t = 4.2 s
a = ?
a = /\v//\t
= (v2 –v1)//\t
= (8.3 m/s – 1.8 m/s)/ 4.2 s
= 1.5 m/s2
If a
= (v2 –v1)//\t, then v2 = v1 + a/\t
and v1 = v2 - a/\t
eg. If a
car starts at 12 m/s, and accelerates 0.62 m/s2 in 15 s, what is the
final speed of the car?
a = 0.62 m/s2
v1 = 12 m/s
/\t = 15 s
v2 = ?
v2 = v1 + a/\t
= 12 m/s + 0.62 m/s2(15 s)
= 12 m/s + 9.3 m/s
= 21.3 m/s
A v/t graph with a steep positive slope is showing quicker
positive acceleration than a graph that is less steep.
A v/t
graph going in a straight flat line (0 slope) is going at constant speed.
A v/t
graph with a negative slope is showing negative acceleration (not deceleration)
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Lesson Eight |
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Summary: (U5) |
Homework: |
ACCELERATION (CONTINUED)
positive
acceleration is speeding up...use a positive sign for acceleration
negative
acceleration is slowing down...use a negative sign for acceleration
no
acceleration occurs at a constant speed...acceleration is 0
Area under a speed/time
graph equals the distance traveled during that time
Average
speed is vav = (v1 + v2)/2 = /\d//\t
Distance travelled is therefor /\d = [(v1 + v2)/2]/\t
If the
speed is constant, the area (distance travelled) under the distance time graph
will be /\d = v/\t = (base)(height) = bh
eg. A dog runs at 15 m/s for 20 seconds. What is the distance the dog runs?
Area (distance traveled) = bh = (15 m/s)(20 s) = 300 m
or Distance = [(v1 + v2)/2]/\t = [(15 m/s + 15 m/s)/2](20s) = [15m/s](20s) = 300 m
If the speed is increasing from a stop the area (distance travelled) under the
distance time graph will /\d = ½ (base)(height) = ½ bh
eg. If
a car accelerates from 0 m/s to 10 m/s in 50 s, then the distance traveled is
Area (distance traveled) = (0.5)bh = (0.5)(10 m/s)(50s)= 250 m
or Distance = [(v1 + v2)/2]/\t = [(0 m/s + 10 m/s)/2](50s) = [5 m/s](50s) = 250 m
If the
speed changes, and the initial speed is greater than 0, the entire area under
the slope must be used to
find
distance traveled. (½ bh + bh)
eg. A
truck accelerates from 5 km/h to 55 km/h in 0.5 h. What is the distance
traveled by the car?
Area
= ½ bh + bh = (0.5)(55km/h - 5km/h)(0.5 h) + (5km/h)(0.5 h)
= (0.5)(50 km/h)(0.5 h) + 2.5 km = 15 km
or Distance = [(v1 + v2)/2]/\t = [(5 km/h + 55 km/h)/2](0.5h) = [30 m/s](0.5h) = 15 km
Do worksheets, using speed/time graphs to find acceleration and distance traveled.
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Lesson Nine |
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Summary: (U6,D8) |
Homework: |
ACCELERATION LAB
-review notes
from last day
-collect
data for acceleration lab
-complete
graphs for acceleration lab
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Lesson Ten |
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Summary: (U3) |
Homework: |
INSTANTANEOUS SPEED
-instantaneous speed is the speed at a particular point on a d vs. t graph
(find the slope of the line at this point)
-if a car goes at a constant speed, the instantaneous speed never changes (find
the slope of the line at any point)
-if
acceleration occurs, the d vs. t graph is curved, and speed changes at different points on the graph.
A TANGENT must be drawn to find the slope at a particular point.
Tangent: a straight line that only touches one point of the
curve, and does not cross or touch the line at any other point.
(practice drawing tangents)
The slope of a tangent at a certain point on a d vs. t (distance vs. time) graph will give the instantaneous speed at that particular point.
On a v vs. t (speed vs. time) graph, the instantaneous speed taken by finding v at the t you are looking at (no slope is needed)
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Lesson Eleven |
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Summary: (D1,D3,D8,R1,R2,R3) |
Homework: |
BALLOON CAR
Do lab to
design and test balloon car for speed and distance.
Repeat
with 3 designs and trials.
Collect
raw data.
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Lesson Twelve |
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Summary: (D1,D3,D5,D8,R1,R2,R3) |
Homework: |
BALLOON CAR CONTINUED
Discuss
development of technology with trial and error.
Show
video on “Collisions”, and discuss car design.
Finish
balloon car lab.
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Lesson Thirteen |
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Summary: (D3,D4,D5,D8) |
Homework: |
SPEED AND ACCELERATION ON AN AIR TABLE
-complete the following lab
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SPEED AND ACCELERATION ON AN AIR TABLE name:________________
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Lesson Fourteen |
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Summary: (U1,U7,U8) |
Homework: |
ACCELERATION AS A VECTOR
a = (v2-v1)//\t
=speed/time (scalar)
a = (v2-v1)//\t
= velocity/time (vector)
eg. An airplane accelerates from a stop to 300 km/h west in 30 s. What is the acceleration of the airplane as a vector in
m/s2?
Assume west is positive
v1
= 0 km/h = 0 m/s
v2
= +300 km/h(1 h/3600 s)(1000 m/1 km) = +83.3 m/s
a
= ?
/\t
= 30 s
a
= (v2-v1)//\t
= (83.3 m/s – 0m/s)/30 s
= 2.8 m/s2
Acceleration
of the plane is 2.8 m/s2 [W]
If initial velocity and acceleration both have the same sign (direction), then the object is speeding up.
If the initial velocity and acceleration have different signs (directions), then the object is slowing down.
eg. The airplane goes from 350 km/h east to a stop in 40 s. What is the
acceleration as the plane slows
down?
Assume
west is positive
v1=
-350 km/h(1 h/3600 s)(1000 m/1 km) = -97.2 m/s
v2 = 0 km/h = 0 m/s
a
= ?
/\t
= 40 s
a =
(v2-v1)//\t
= (0 m/s –(-97.2 m/s))/40 s
= +2.4 m/s2
Acceleration
of the plane is 2.4 m/s2 [W]
If a ball is thrown up, its acceleration is in downwards direction, and this acceleration will stay the same, even when the speed of the ball changes directions.
eg. A ball is thrown up at 10 m/s, while the acceleration will be 9.81 m/s2
down. After 0.5 s what is the velocity
of the ball.
Assume up is positive
v1
= +10 m/s
v2=
?
a
= -9.8 m/s2
/\t
= 0.5 s
If a = (v2-v1)//\t,
then v2 = v1 + a/\t
v2 = v1
+ a/\t
= 10 m/s + (-9.8 m/s2)(0.5 s)
= 5.1 m/s
Final
velocity of the ball is 5.1 m/s up
eg. After 5
s, what is the velocity of the same ball?
Assume up is positive
v1
= +10 m/s
v2=
?
a
= -9.8 m/s2
/\t
= 5 s
If a = (v2-v1)//\t,
then v2 = v1 + a/\t
v2 = v1 + a/\t
= 10 m/s + (-9.8 m/s2)(5 s)
= -39.0 m/s
Final
velocity of the ball is 39.0 m/s down
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Lesson Fifteen |
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Summary: |
Homework: |
TEST REVIEW
Review for test on lessons 1-13
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Lesson Sixteen |
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Summary:(U7,U8) |
Homework: |
FINDING DISPLACEMENT
Displacement
can be determined by finding the area under a velocity vs. time graph.
If vav = /\d//\t then /\d = vav/\t
Problem: A car is going 2.5 m/s east in 3 s, and then 5 m/s west in 4 s.
What is the resultant
displacement of the car?
Assume
east is positive
vav1 = 2.5 m/s
/\t1
=
3 s
/\d1 = ?
vav2 = -5 m/s
/\t2
=
4 s
/\d2 = ?
/\dR = ?
/\dR = /\d1 + /\d2
= (vav1/\t1)
+ (vav2/\t2)
= (2.5 m/s)(3 s) + (-5 m/s)(4s)
= 7.5 m – 20 m
= -12.5 m
= 12.5 m [W]
The formula for /\d = vav/\t works when the velocity is constant
[the area under the graph is length (vav) X width (/\t), which is the area of
a rectangle].
Does it work if the velocity changes, due to acceleration?
A rabbit from 0.0 m/s west to 8.0 m/s west over the course of 6.0 s.
This
will form a triangular graph. The area under this graph is /\d = (0.5)/\v/\t
As the average velocity is actually vav = /\v/2
= (v1 + v2)/2
(v2+ v1)/2 = /\d//\t can be rearranged to solve for displacement
/\d = /\t[(v1 + v2)/2] is the
formula that should always be used
Assume west is
positive.
/\d = vav/\t
= /\t[(v1 + v2)/2]
= (6.0 s)(0.0 m/s + 8.0 m/s)/2
= 24 m
= 24 m [W]