Download Lecture 5 Friday Sept 5

Physics 31
Lecture 5:
Motion in One Dimension
With Changing Velocity
v(t) from x(t)
•
x
vx (t ) 
average velocity
t
in the time interval t
v(t) is slope of x(t)
dx(t )
v(t ) 
instantaneous velocity
dt
x(t) from v(t)
v(t)
x  vx t
tend
What does x(t) look like?
t
v(t) = constant=10m/s
1. How far would the object travel
in one second?
2. How far in two seconds?
Slope is v = constant
x(t)
t
Acceleration
Acceleration is:
•
The rate of change of
velocity
•
The slope of a velocityversus-time graph
Slide 2-20
Checking Understanding
These four motion diagrams show the motion of a particle along
the x-axis. Rank these motion diagrams by the magnitude of the
acceleration. There may be ties.
Slide 2-21
Checking Understanding
These four motion diagrams show the motion of a particle along
the
x-axis. Which motion diagrams correspond to a positive
acceleration? Which motion diagrams correspond to a negative
acceleration?
Slide 2-22
Checking Understanding
These six motion diagrams show the motion of a particle along
the
x-axis. Rank the accelerations corresponding to these motion
diagrams, from most positive to most negative. There may be
ties.
Slide 2-23
Acceleration—Rate of Change of
Velocity
v x
ax (t ) 
average rate of change of velocity
t
in the time t = average acceleration
dv (t )
ax (t ) 
= slope of v(t) versus t =
dt
instantaneous acceleration
v(t) from a(t)
a(t )  constant  a
tf
v(t )  v(ti )   a dt  v(ti )  at f  ati 
ti
v(ti )  a t
vx (t )  v0 x  axt
x(t) from v(t)
x(t) is the area under the v(t) vs t curve
For constant a
vx (t )  vix  ax t
area under vx curve is:
tf
x(t )  xi   (vx (t ))dt 
ti
xi  
tf
ti
1 2
(vix  axt ) dt  xi  vixt f  axt f
2
In general, for constant acceleration
1 2
x(t )  xi  voxt  axt
2
Quiz, question 1
Here is a motion diagram of a car moving along a straight stretch
of road:
Which of the following velocity-versus-time graphs matches this
motion diagram?
A.
B.
C.
D.
Slide 2-13
Question 2
A graph of position versus time for a
basketball player moving down the
court appears like so:
Which of the following velocity graphs matches the above
position graph?
A.
B.
C.
D.
Slide 2-15
Question 3
• A car is moving with a constant velocity in the
positive x-direction along a straight road. The
magnitude of the velocity is v=12 m/s. Ten
seconds later, it crashes into a wall and stops
very suddenly.
• A) sketch v(t) versus t for the first 15 seconds.
• B) sketch x(t) versus t for the first 15 seconds.
Homework due Monday
Reading 2: 3-4 Motion with constant
acceleration
Problems: 2: CQ1, CQ6, MC18, 1, 3, 5, 8, 12