Slide 1 : RECTILINEAR KINEMATICS: ERRATIC MOTION Today’s Objectives:
Students will be able to:
Determine position, velocity, and acceleration of a particle using graphs. In-Class Activities:
Check Homework
Reading Quiz
Applications
s-t, v-t, a-t, v-s, and a-s diagrams
Concept Quiz
Group Problem Solving
Attention Quiz
Slide 2 : READING QUIZ 1. The slope of a v-t graph at any instant represents instantaneous
A) velocity. B) acceleration.
C) position. D) jerk. 2. Displacement of a particle in a given time interval equals the area under the ___ graph during that time.
A) a-t B) a-s
C) v-t C) s-t
Slide 3 : APPLICATION In many experiments, a velocity versus position (v-s) profile is obtained.
If we have a v-s graph for the tank truck, how can we determine its acceleration at position s = 1500 feet?
Slide 4 : ERRATIC MOTION
(Section 12.3) The approach builds on the facts that slope and differentiation are linked and that integration can be thought of as finding the area under a curve.
Slide 5 : S-T GRAPH
Slide 6 : V-T GRAPH Also, the distance moved (displacement) of the particle is the area under the v-t graph during time ?t.
Slide 7 : A-T GRAPH
Slide 8 : A-S GRAPH
Slide 9 : V-S GRAPH Another complex case is presented by the velocity vs. distance or v-s graph. By reading the velocity v at a point on the curve and multiplying it by the slope of the curve (dv/ds) at this same point, we can obtain the acceleration at that point. Recall the formula
a = v (dv/ds).
Thus, we can obtain an a-s plot from the v-s curve.
Slide 10 : EXAMPLE What is your plan of attack for the problem?
Slide 11 : EXAMPLE (continued) Solution: The v-t graph can be constructed by finding the slope of the s-t graph at key points. What are those?
when 0 < t < 5 s; v0-5 = ds/dt = d(3t2)/dt = 6 t m/s
when 5 < t < 10 s; v5-10 = ds/dt = d(30t?75)/dt = 30 m/s
v-t graph
Slide 12 : EXAMPLE (continued) Similarly, the a-t graph can be constructed by finding the slope at various points along the v-t graph.
when 0 < t < 5 s; a0-5 = dv/dt = d(6t)/dt = 6 m/s2
when 5 < t < 10 s; a5-10 = dv/dt = d(30)/dt = 0 m/s2
a-t graph
Slide 13 : CONCEPT QUIZ 1. If a particle starts from rest and
accelerates according to the graph
shown, the particle’s velocity at
t = 20 s is
A) 200 m/s B) 100 m/s
C) 0 D) 20 m/s 2. The particle in Problem 1 stops moving at t = _______.
A) 10 s B) 20 s
C) 30 s D) 40 s
Slide 14 : GROUP PROBLEM SOLVING Find slopes of the v-t curve and draw the a-t graph.
Find the area under the curve. It is the distance traveled. Finally, calculate average speed (using basic definitions!). Given: The v-t graph shown.
Find: The a-t graph, average speed, and distance traveled for the 0 - 90 s interval.
Plan:
Slide 15 : GROUP PROBLEM SOLVING
(continued) Solution: Find the a–t graph:
For 0 = t = 30 a = dv/dt = 1.0 m/s²
For 30 = t = 90 a = dv/dt = -0.5 m/s²
a-t graph
Slide 16 : GROUP PROBLEM SOLVING (continued) Now find the distance traveled:
Ds0-30 = ò v dt = (1/2) (30)2 = 450 m
Ds30-90 = ò v dt
= (1/2) (-0.5)(90)2 + 45(90) – (1/2) (-0.5)(30)2 – 45(30)
= 900 m
s0-90 = 450 + 900 = 1350 m
vavg(0-90) = total distance / time
= 1350 / 90
= 15 m/s
Slide 17 : ATTENTION QUIZ 1. If a car has the velocity curve shown, determine the time t necessary for the car to travel 100 meters.
A) 8 s B) 4 s
C) 10 s D) 6 s
Slide 18 :