Phys101 Lecture 2 - 1
Phys101 Lecture 2
Kinematics in One Dimension
Key points:
• Always set up a coordinate system first (why?)
• Position, displacement, velocity and acceleration
• Motion at constant acceleration
• Calculus as a convenient tool
You are expected to know everything in sections 2-1,2,3,4,5,6,7.
Phys101 Lecture 2 - 2
Example 2-20: Ball thrown upward at edge of cliff.
A ball is thrown upward at a speed of 15.0 m/s by a person
standing on the edge of a cliff. How long does it take for the ball to
reach the base that is 50.0m below the cliff?
i>Clicker question (2-1): Which coordinate system can be used?
A.
B.
C.
D.
E.
y
B
o
o
A
A only
A or C
A or B
C only
A or B or C or D
y
y
D
o
o
C
y
Phys101 Lecture 2 - 3
Position, Displacement and Distance
Example:
John walked 70 m east,
then 30 m west. It took John 50
seconds to complete the walk.
• Position: x (when t=0, x=0; when t=50 s, x=40 m)
• Displacement: Δx=x2–x1=40–0=40m (change in position)
Δx represents how far the object is from its starting point, regardless
of how it got there (blue line).
• Distance (dashed line) is the length measured along the actual path.
Distance = 70 + 30 = 100m
Displacement can have different signs (+/-);
but distance is always positive.
Phys101 Lecture 2 - 4
Average Speed and Average Velocity
In the previous example,
100m
average speed 
 2.0m / s
50s
average velocity 
40m
 0.80m / s
50s
Phys101 Lecture 2 - 5
Instantaneous Velocity – a better way to
describe how fast an object is moving
The instantaneous velocity is the average velocity in the
limit as the time interval becomes infinitesimally short.
dx
dt
-- derivative of x with respect to t
t – independent variable
x – dependent variable
i.e., x is a function of t: x = f(t)
dx
Meaning of
: slope of tangent of curve x  f (t )
dt
Phys101 Lecture 2 - 6
Instantaneous Velocity
On a graph of a particle’s position vs. time, the instantaneous velocity is the
slope of the straight line tangent to the x-t curve. (The average velocity is the
slope of the secant line.)
Average velocity
x
v
 slope of secant P1P2
t
Instantaneous velocity
dx
v
 slope of tangent at P1
dt
Phys101 Lecture 2 - 7
The derivative of polynomial functions
• The power rule:
If x  f (t )  t n , then
d (t n )
i.e.,
 nt n 1.
dt
dx
 nt n 1.
dt
   nt
Can also be written as t n
n 1
, or f (t )  nt n 1.


dx
 2t. or simply (t 2 )  2t
dt
dx
e.g., if x  t , then
 1. (the slope is constant)
dt
dx
e.g., if x  1, then
 0. (the derivative of a constant is 0)
dt
e.g., if x  t 2 , then
• The polynomial function:
If x  f (t )  a0  a1t  a2t 2  a3t 3 ,
then


dx
d

a0  a1t  a2t 2  a3t 3
dt
dt
 ( a0 )  ( a1t )  ( a2t 2 )  ( a3t 3 )
 0  a1 (t )  a2 (t 2 )  a3 (t 3 )
 a1  2a2t  3a3t 2
Phys101 Lecture 2 - 8
Example 2-3: Given x as a function of t, find the
instantaneous velocity
A jet engine moves along an experimental
track (which we call the x axis) as shown.
Its position as a function of time is given
by the equation x = At2 + B, where A =
2.10 m/s2 and B = 2.80 m. Determine the
instantaneous velocity at t = 5.00 s.
Phys101 Lecture 2 - 9
Acceleration
Acceleration is the rate of change of velocity.
average accelerati on  a 
v v2  v1

t t2  t1
dv
v
instantaneous accelerati on  a 
 lim
dt t 0 t
Instantaneous acceleration is
the derivative of v with
respect to t.
i.e., a is the slope of the v-t
curve.
Demo: x-t, v-t and a-t curves
Of a falling basketball
Phys101 Lecture 2 - 10
Example 2-7: Given x(t), find the acceleration
A particle is moving in a straight line so that its position is given by the
relation x = (2.10 m/s2)t2 + (2.80 m). Calculate (a) its average
acceleration during the time interval from t1 = 3.00 s to t2 = 5.00 s, and
(b) its instantaneous acceleration as a function of time.
Phys101 Lecture 2 - 11
Motion at Constant Acceleration
Useful formulas (can you derive them?)
Assumptions
• constant acceleration
• when t=0, x=x0, v=v0
i>Clicker question (2-2):
Can the formulas be used when the acceleration varies with time?
A. Yes.
B. No.
C. It depends.
Phys101 Lecture 2 - 12
Example 2-20: Ball thrown upward at edge of cliff.
A ball is thrown upward at a speed of 15 m/s by a person standing on
the edge of a cliff. How long does it take for the ball to reach the base
that is 50m below the cliff?
i>Clicker question (2-3): For coordinate system D,
y
B
o
o
A
A.
B.
C.
D.
E.
y0=50m, v0=15m/s, a=9.8m/s2
y0=-50m, v0=15m/s, a=9.8m/s2
y0=-50m, v0=-15m/s, a=9.8m/s2
y0=-50m, v0=-15m/s, a=-9.8m/s2
y0=0, v0=-15m/s, a=9.8m/s2
y
y
D
o
o
C
y
Phys101 Lecture 2 - 13
Next: Vectors and 2-d Kinematics
• Read 3-1,2,3,4,5 before coming to the lecture
(see the course calendar for more info).
• Print out and read the lecture note
(lecture03A.pdf) and be ready for clicker
quizzes.
• Attempt assignment #1 (both MP and written)
and bring your questions to the tutorial Monday.