UNIT 1: 1-D KINEMATICS Lesson 4:

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UNIT 1: 1-D KINEMATICS Lesson 4:
CENTRE HIGH: PHYSICS 20 UNIT 1: 1-D KINEMATICS Lesson 4: Graphical Analysis of Accelerating Motion

Reading Segment #1: Average and Instantaneous Velocity on Position-Time Graphs To prepare for this section, please read: Unit 1: p.14

D1. Position-Time Graphs (Accelerated Motion)
Recall, when the motion of an object is uniform, - the velocity is constant - since velocity is the slope of a position-time graph, the graph is a straight line (constant slope). But now, the velocity is going to change. As a result, the line is no longer going to be straight.

Consider the following position-time graph
for an accelerating object: (Forward is positive) d (m) t (s) Can you describe the motion?

d (m) Zero slope t (s) At the start, the slope is zero. Thus, the object starts from rest.

d (m) Slopes are increasing t (s) But, over time, the slope increases. Thus, the object's velocity increases, which in this case means that the object is speeding up.

d (m) Slopes are increasing t (s) Since the velocity is changing, the object is accelerating.

For non-linear position-time graphs, there are two types
of velocities (i.e. slopes) that could be calculated: 1. Average velocity - from one time to another 2. Instantaneous velocity - at one point in time

1. Average velocity To find the average velocity from t = t1 to t = t2,  Find the initial position (at t1) and the final position (at t2) on the graph  Draw a straight line from the two positions (called a secant line)  Find the slope of this secant line

e.g. How would we find the average velocity from t = 3.0 s
to t = 5.0 s on the graph below? d (m) t (s)

d (m) t (s) First, find the initial and final position on the line.

d (m) secant line t (s) Draw a straight line between the two points. This is called the secant line.

d (m) slope = vavg t (s) Find the slope of the secant line. The slope is the average velocity between 3.0 and 5.0 s.

2. Instantaneous velocity
To find the instantaneous velocity at t = t1 :  Find the position (at t1) on the graph  Draw a tangent line at this point  Choose 2 points on the tangent line and find its slope

e.g. How would we find the instantaneous velocity at t = 4.0 s ?
d (m) 4.0 t (s)

d (m) 4.0 t (s) First, find the position at t = 4.0 s on the line.

d (m) tangent line 4.0 t (s) Next, draw a tangent line at this point. A tangent line touches the curve only at this point, and if done correctly, it should show the slope at that point.

d (m) slope of tangent = velocity 4.0 t (s) Finally, locate two points on the tangent line and find its slope. The slope is the instantaneous velocity.

Ex. 1 For the graph below, find: (North is positive)
a) the average velocity from 0.25 s to 5.0 s b) the instantaneous velocity at 1.0 s

a) Locate the positions at t = 0.25 s and t = 5.0 s

secant line a) Draw a straight line between the two points. This the secant line.

(5.0 s, 50.0 m) secant line (0.25 s, 11.0 m) a) Choose two points on the secant line and find the slope.

a) Average velocity between 0.25 s and 5.0 s:
vavg = slope of secant line = y2 - y1 = (50.0 m) - (11.0 m) x2 - x1 (5.0 s) - (0.25 s) = m/s North Be sure to include the direction for velocity.

b) Locate the position at t = 1.0 s

tangent line b) Draw the tangent line at this point. It should touch the curve only at this point.

(3.25 s, 48.0 m) tangent line (1.0 s, 22.0 m) b) Find two points on the tangent line. Calculate the slope.

b) Instantaneous velocity between 1.0 s:
v1.0 = slope of tangent line = y2 - y1 = (48.0 m) - (22.0 m) x2 - x1 (3.25 s) - (1.0 s) = m/s North Be sure to include the direction for velocity.

Practice Problems Try these problems in the Physics 20 Workbook: Unit 1 p. 16 #1

D2. Velocity-Time Graphs (Accelerated Motion)
There are two major calculations we find for velocity-time graphs: 1. Acceleration - this will be the slope of a v-t graph 2. Displacement - this will be the area "under" the v-t graph

1. Acceleration on a velocity-time graph
The slope of a v-t graph is the acceleration Thus, if the object has uniform acceleration: - the acceleration is constant - the slope is constant on the v-t graph - this produces a straight line

e.g. Ref: Forward is positive
Backward is negative If the graph has a positive slope (i.e. a positive acceleration), it means the object's velocity is increasing. v v t

Ref: Forward is positive
Backward is negative If the graph has a negative slope (i.e. a negative acceleration), it means the object's velocity is decreasing. v v t

Ref: Forward is positive
Backward is negative If the graph has a zero slope (i.e. a zero acceleration), it means the object has a constant velocity. v v t

Ex. 2 For the graph below, find the acceleration at t = 3.0 s.
(East is positive) Velocity 8.0 ( 103 m/s) 2.0 4.0 t (s)

Velocity 8.0 ( 103 m/s) 2.0 4.0 t (s) Acceleration is the slope of a v-t graph. Since this is a straight line, you can choose any 2 points on the line to calculate the slope.

Velocity (0 s, 8.0  103 m/s) ( 103 m/s) (4.0 s, 2.0  103 m/s) 2.0 4.0 t (s) Choose 2 points on the line.

Acceleration at 3.0 s: a3.0 = slope of line = y2 - y1 = (2.0  103 m/s) - (8.0  103 m/s) x2 - x (3.0 s) - (0 s) =  103 m/s2 =  103 m/s2 West Be sure to include the direction for acceleration.

Ex. 3 Describe the motion showed by the following
velocity-time graph: (Forward is positive) v (m/s) 10 t (s) -6

v (m/s) Forward is positive
t (s) -6 At the very start of the motion (A), the object is moving forward at 10 m/s.

v (m/s) Forward is positive
B t (s) -6 Along AB, the object is slowing down at a constant acceleration (constant slope), from 10 m/s to 0. It is still moving forward, since the velocity is positive. At B, the object is at rest.

v (m/s) Forward is positive
B t (s) C Along BC, the object is speeding up at a constant acceleration (constant slope), from 0 to 6 m/s. It is now moving backward, since the velocity is negative. At C, the object is at moving backwards at 6 m/s.

v (m/s) Forward is positive
B t (s) C D Along CD, the object is at a constant velocity of -6.0 m/s. Acceleration (slope) is zero. At D, the object is at moving backwards at 6 m/s.

v (m/s) Forward is positive
B E t (s) C D Along DE, the object is slowing down at a constant acceleration (slope), from 6 m/s to zero. It is moving backwards, since the velocities are negative. At E, the object has come to rest.

2. Displacement on a velocity-time graph
The area between the line and the t-axis on a v-t graph is the displacement

Ref: Forward is positive
Backward is negative e.g. When the line is above the t-axis, displacement (area) is positive. i.e. A forward displacement v v t t

Ref: Forward is positive
Backward is negative e.g. When the line is below the t-axis, displacement (area) is negative. i.e. A backward displacement v v t t

Ex. 4 Calculate the displacement for the first 8.0 seconds.
(North is positive) v (m/s) 33 14 8.0 t (s)

v (m/s) 33 14 8.0 t (s) Displacement is the area between the line and the t-axis

v (m/s) 33 1 14 2 8.0 t (s) To calculate the area, find the area of the triangle (1) and the rectangle (2).

v (m/s) 33 m/s 14 m/s 8.0 s t (s) Show the dimensions on the diagram.

Displacement for the first 8.0 seconds:
A1 = Area of triangle = b h = (8.0 s) (19 m/s) = m A2 = Area of rectangle = L  w = (8.0 s) (14 m/s) = 112 m Thus, d = A1 + A2 =  m North

Practice Problems Try these problems in the Physics 20 Workbook: Unit 1 pp #2 - 5