Chapter 3. Acceleration and Newton’s Second Law of Motion. Mechanics (Classical/Newtonian Mechanics): Study of motion in relation to force and energy, ie, the effects of force and energy on the motion of an object.
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Mechanics (Classical/Newtonian Mechanics): Study of motion in relation to force and energy, ie, the effects of force and energy on the motion of an object.
Kinematics – Mechanics that describes how objects move, no reference to the force or mass.
Dynamics– Mechanics that deals with force: why objects move.
Reference frames and coordinates
When we say a car is moving at 45 km/hr, we usually mean “with respect to the earth” although is is not explicitly stated.
An escalator is moving at 3 m/s relative to the ground. A lady walks on the escalator at 2 m/s relative to the escalator. The lady’s speed relative to the ground is 5 m/s.
1D coordinate system:
• Specify the origin.
• Show where x axis is pointing: to the right is positive direction of x.
• Label the axis with the relevant units.
y

+
0
x

Reference Direction:2D coordinate system:
• Specify the origin.
• Show x and y axes. To the right is positive direction of x. Vertically upward is positive y.
• Label the axis with the relevant units.
3
A
x
3
3
B
3
Position
10
The displacement from A to B is
xdirection: 1 – 3 = 4
ydirection: 2 – 1 = 3
y
3
A
x
3
3
B
3
13
30 km E
40 km N
Average Speed and Average Velocity What is the magnitude of its
Average Speed = (distance traveled)/time taken = d/t
Speed:  specified only by magnitude. It is a scalar quantity.
SI unit = m/s
 always a positive number
Average velocity = (displacement)/time taken
= (x2 – x1)/(t2 – t1) = x/t
Velocity:  specified by both magnitude and direction.
 It is a vector quantity.
 units – m/s
 Positive/ negative sign used to indicate
direction
Magnitude of average speed and average velocity: not always equal.
Charles walks 120 m due north in 40s. He then walks another 60 m still due north in another 60s.
Average speed?
Average velocity?
Average speed = Total distance/time
Average velocity = Displacement/time
Charles walks 120 m due north. He then walks another 60 m due south. He took a total of 100 s for the journey.
Average speed?
Average velocity?
Average speed = Total distance/time
Average velocity = Displacement/time
Charles walks 120 m due north. He then walks another 120 m due south. He took a total of 100 s for the journey.
Average speed?
Average velocity?
Instantaneous Velocity due south. He took a total of 100 s for the journey.
x(t)
Dx
Dt
t
x(t)
t
23
Instantaneous Velocity due south. He took a total of 100 s for the journey.
Instantaneous velocity = Average speed.
uniform (constant) velocity
velocity (m/s)
velocity (m/s)
t2
t1
time (s)
time (s)
Determine the velocity of the car at times A, B, C and D. due south. He took a total of 100 s for the journey.
A Positive Zero Negative
B Positive Zero Negative
C Positive Zero Negative
D Positive Zero Negative
x(t)
A
t
B
C
D
v(t)
x(t)
t
t
v = slope
A car moves at a constant velocity of magnitude 20 m/s. At time t = 0, its position is 50 m from a reference point. What is its position at
(i) t = 2s? (ii) t = 4s (iii) t = 10s?
x(t)
v(t)
t
t
Area Under velocitytime Graph time t = 0, its position is 50 m from a reference point. What is its position at
Area under the velocitytime graph = magnitude of the displacement over the time interval.
uniform (constant) velocity
velocity (m/s)
Area = v(t2t1) = (x2 – x1) = x
time (s)
t2
t1
velocity (m/s)
time (s)
A stray dog ran 3 km due north and then 4 km due east. What is the magnitude of its average velocity if it took 45 min?
v(t) is the magnitude of its average velocity if it took 45 min?
D
C
Dv
B
E
A
Dt
t
Acceleration (a)
v(t) is the magnitude of its average velocity if it took 45 min?
t
Acceleration (a)
a(t)
v(t)
v
t
t
t
a = slope
A car moves at a constant acceleration of magnitude 5 m/s is the magnitude of its average velocity if it took 45 min?2. At time t = 0, the magnitude of its velocity is 8 m/s. What is the magnitude of its velocity at
(i) t = 2s? (ii) t = 4s? (iii) t = 10s?
Acceleration (a) is the magnitude of its average velocity if it took 45 min?
Average acceleration =v/ t
T/F? If the acceleration of an object is zero, it must be at rest.
T/F? If an object is at rest, its acceleration must be zero.
Example is the magnitude of its average velocity if it took 45 min?
Train A moves due east along a straight line with a velocity 8 m/s. Within 7 seconds, it’s velocity increases to 22 m/s. What is its average acceleration?
Train B is moves due east along a straight line at a velocity of 18 m/s. Within 10 s, its velocity drops to 3 m/s. What is its average acceleration?
When an object slows down, we say it is decelerating. In that case, the direction of the acceleration will be opposite to that of the velocity.
Example is the magnitude of its average velocity if it took 45 min?
Is it possible for an object to have a positive velocity at the same time as it has a negative acceleration?
1  Yes
2  No
“Yes, because an object can be going forward but at the same time slowing down which would give it a negative acceleration.”
37
Graphical Representation is the magnitude of its average velocity if it took 45 min?
Slope of the line = average velocity
Position, x (m)
Time, t (s)
If the slope is zero, the object is at rest
Graphical Representation is the magnitude of its average velocity if it took 45 min?
Slope of the line = average acceleration
velocity, v (m/s)
O
Time, t (s)
If the slope is zero, the object is moving with zero acceleration (constant velocity)
velocity, v (m/s) is the magnitude of its average velocity if it took 45 min?
C
D
E
B
A
O
Time, t (s)
Slope of the line = average acceleration = v/t
1. When is a = 0?
2. When is a < 0?
3. When is a = maximum?
4. A train moves due north along a straight path with a uniform acceleration of 0.18 m/s2. Ifits velocity is 2.4 m/s, what will its velocity be after 1 minute?
6. A bird flew 6.00 km north, then turned around and flew 1.50 km west. If it took 25.0 min, what was magnitude of the average velocity of the bird?
7. A train moving along a straight path due north has a velocity of 20 m/s. Within 5.0 seconds,its velocity became 5.0 m/s. What was the train's average acceleration?
VECTORS velocity of 20 m/s. Within 5.0 seconds,
tip
Vector A
tail
+y
A
Equal Vectors:
A = B only if they have equal magnitudes and same directions.
B
+x
Displacing a vector parallel to itself does not change it
The negative of a vector: velocity of 20 m/s. Within 5.0 seconds,
Two vectors with equal magnitude but opposite directions are negatives of each other
A
A
Vectors can be multiplied by a scalar:
A
2A
½ A
Components of a vector velocity of 20 m/s. Within 5.0 seconds,
A vector can be expressed as a sum of 2 vectors called “components” that are parallel to the x and y axes:
+y
A = Ax + Ay
Ax = x component of A.
Ay = ycomponent of A.
A
Ay
+x
Ax
h = hypotenus velocity of 20 m/s. Within 5.0 seconds,
o = opposite
a = adjacent
SOH CAH TOA
sin = opposite/hypotenuse = o/h
cos = adjacent/hypotenuse = a/h
tan = opposite/adjacent = o/a
Pythagorean theorem: h2 = a2 + o2
A velocity of 20 m/s. Within 5.0 seconds,
Ay
Ax
sin = o/h = Ay/A or ycomponent:Ay = A . sin
cos = a/h = Ax/A or xcomponent: Ax = A . cos
A2 = Ax2 + Ay2 or magnitude of vectorA = (Ax2 + Ay2 )
tan = o/a = Ay/Ax or = tan1(Ay/Ax)
Addition of vectors velocity of 20 m/s. Within 5.0 seconds,
Three methods:
Component method (analytical).
Tiptotail (graphical).
Parallelogram (graphical).
Component method velocity of 20 m/s. Within 5.0 seconds,
Add vectors V1 + V2 + V3:
{V1=V1x + V1y}, {V2 = V2x + V2y}, {V3 = V3x + V3y
2. Add xcomponents and ycomponents separately:
{Vx = V1x + V2x +V3x} and {Vy = V1y + V2y +V3y}
3. Find the magnitude of the resultant vector using pythagorean theorem: V = {V2x + V2y}
4. Find the angle of the resultant measured from the +x axis: = tan1(Vy/Vx)
Example velocity of 20 m/s. Within 5.0 seconds,
y
A
B
C
30o
x
Three displacement vectors are shown in the figure below. Their magnitudes are A = 20 cm, B = 16 cm and C = 12 cm. Find the magnitude and angle of the resultant vector.
Tiptotail Method velocity of 20 m/s. Within 5.0 seconds,
Draw the vectors such that the “tail” of the second vector connects to the tip of the first vector. The resultant is from the tail of the first to the tip of the second.
A
B
B
C
A
B
A
D
C = A + B
D = A  B
Parallelogram Method velocity of 20 m/s. Within 5.0 seconds,
Draw the vectors such that their “tails” are joined to a common origin.
Construct a parallelogram with the two vectors as adjacent sides.
The resultant vector is the diagonal line of the parallelogram drawn from the common origin.
A
B
C = A + B
C
A
B
Z vector is2
Y
Z1
X
From the diagram shown in the figure below, what are vectors Z1 and Z2 in terms of X and Y?
In the diagram below, what are the x and ycomponents of vector A if the magnitude of A is 40 units?
y
A
35o
x
Z vector A if the magnitude of A is 40 units? 2
A
A
Z1
B
B
From the diagram shown in the figure below, what are vectors Z1 and Z2 in terms of B and A?
Relative Velocity vector A if the magnitude of A is 40 units?
You are on a train traveling 40 mph North. If you walk 5 mph toward the front of the train, what is your speed relative to the ground?
1) 45 mph 2) 40 mph 3) 35 mph
40 mph N + 5 mph N = 45 mph N
40
5
45
25
Relative Velocity vector A if the magnitude of A is 40 units?
You are on a train traveling 40 mph North. If you walk 5 mph toward the rear of the train, what is your speed relative to the ground?
1) 45 mph 2) 40 mph 3) 35 mph
40 mph N  5 mph N = 35 mph N
40
5
35
28
Relative Velocity vector A if the magnitude of A is 40 units?
You are on a train traveling 40 mph North. If you walk 5 mph sideways across the car, what is your speed relative to the ground?
1) < 40 mph 2) 40 mph 3) >40 mph
40 mph N + 5 mph W = 41 mph N
5
40
Relative Motion (Add vector components)
30
P2.27: Find the magnitude and direction of the vector with the following components:
P2.56: A 3.0 kg block is at rest on a horizontal floor. If you push horizontally on the block with a force of 12.0 N, it just starts to move.
(b) A 7.0kg block is stacked on top of the 3.0kg block. What is the magnitude F of the force acting horizontally on the 3.0kg block as before, that is required to make the two blocks start to move?
P3.47: A 2010kg elevator moves with an upward acceleration of 1.50 m/s2. What is the tension that supports the elevator?
P3.48: A 2010kg elevator moves with a downward acceleration of 1.50 m/s2. What is the tension that supports the elevator?
A stray dog ran 4 km due north and then 4 km due east. What is the magnitude of its displacement after this movement?