1-D kinematics...


Consider limit t1 t2
Instantaneous velocity v is defined as:
v( t ) 
x
x
dx( t )
dt
so v(t2) = slope of line tangent to path at t2.
x
2
x
1
t1
t2
t
t
Physics 211: Lecture 1, Pg 1
1-D kinematics...


Acceleration a is the “rate of change of velocity”
Average acceleration aav in the time t = t2 - t1 is:
aav 

v ( t 2 )  v ( t1 ) v

t 2  t1
t
And instantaneous acceleration a is defined as:
dv ( t ) d 2 x( t )
a( t ) 

dt
dt 2
using v ( t ) 
dx( t )
dt
Physics 211: Lecture 1, Pg 2
Recap

If the position x is known as a function of time, then we can
find both velocity v and acceleration a as a function of time!
x
x  x( t )
dx
dt
dv
d 2x
a 

dt
dt 2
v 
v
a
t
t
t
Physics 211: Lecture 1, Pg 3
More 1-D kinematics


We saw that v = dx / dt
In “calculus” language we would write dx = v dt, which we
can integrate to obtain:
t2
x (t 2 )  x (t1 )   v (t )dt
t1

Graphically, this is adding up lots of small rectangles:
v(t)
+ +...+
= displacement
t
Physics 211: Lecture 1, Pg 4
1-D Motion with constant acceleration


1
t n 1  const
n 1
dv
Also recall that a 
dt
n
Math 220:  t dt 

If a is constant, we can integrate this using the above rule
to find:
v   a dt  a  dt  at  v 0

Similarly, since v 
dx
we can integrate again to get:
dt
1
x   v dt   ( at  v 0 )dt  at 2  v 0 t  x0
2
Physics 211: Lecture 1, Pg 5
Recap

So for constant acceleration we find:
Ramp
w/ lights
x
1
x  x0  v 0 t  at 2
2
v  v 0  at
a  const
v
a
t
t
t
Physics 211: Lecture 1, Pg 6
Lecture 1, Act 2
Motion in One Dimension

When throwing a ball straight up, which of the following is
true about its velocity v and its acceleration a at the
highest point in its path?
(a) Both v = 0 and a = 0.
(b) v  0, but a = 0.
y
(c) v = 0, but a  0.
Physics 211: Lecture 1, Pg 7
Lecture 1, Act 2
Solution

Going up the ball has positive velocity, while coming down
it has negative velocity. At the top the velocity is
momentarily zero.
x

Since the velocity is
continually changing there must
v
be some acceleration.
 In fact the acceleration is caused
by gravity (g = 9.81 m/s2).
 (more on gravity in a few lectures) a

The answer is (c) v = 0, but a  0.
t
t
t
Physics 211: Lecture 1, Pg 8
Useful Formula: 1-D motion with constant
acceleration
dv d 2 x

a
dt
dt


1
x  x0  v 0 t  at 2
2
v  v0  at
Solving for t:
Plugging in for t:
t
v  v0
a
 v  v0  1  v  v0 
x  x0  v 0 
  a

 a  2  a 
2
v 2  v 0  2a( x  x0 )
2
Physics 211: Lecture 1, Pg 9
Recap:

For constant acceleration:
Washers
1
x  x0  v 0 t  at 2
2
v  v 0  at
a  const

From which we know:
v 2  v 02  2a(x  x0 )
v av
1
 (v 0  v)
2
Physics 211: Lecture 1, Pg 10
Problem 1

A car is traveling with an initial velocity v0. At t = 0, the
driver puts on the brakes, which slows the car at a rate of
ab
vo
ab
x = 0, t = 0
Physics 211: Lecture 1, Pg 11
Problem 1...

A car is traveling with an initial velocity v0. At t = 0, the
driver puts on the brakes, which slows the car at a rate of
ab. At what time tf does the car stop, and how much farther
xf does it travel?
v0
ab
x = 0, t = 0
v=0
x = xf , t = tf
Physics 211: Lecture 1, Pg 12
Problem 1...

Above, we derived: v = v0 + at

Realize that a = -ab

Also realizing that v = 0 at t = tf :
find 0 = v0 - ab tf or
x
tf = v0 /ab
Physics 211: Lecture 1, Pg 13
Problem 1...

To find stopping distance we use:
v 2  v 02  2a(x  x0 )

In this case v = vf = 0, x0 = 0 and x = xf
 v 0  2( ab )xf
2
2
v
xf  0
2 ab
Physics 211: Lecture 1, Pg 14
Problem 1...



So we found that
2
1 v0
tf 
, xf 
ab
2 ab
v0
Suppose that vo = 65 mi/hr = 29 m/s
Suppose also that ab = g = 9.81 m/s2
 Find that tf = 3 s and xf = 43 m
Physics 211: Lecture 1, Pg 15
Problem Solving Tips:




Read Carefully!
 Before you start work on a problem, read the problem
statement thoroughly. Make sure you understand what
information is given, what is asked for, and the meaning of
all the terms used in stating the problem.
Using what you are given, set up the algebra for the problem and
solve for your answer algebraically
 Invent symbols for quantities you know as needed
 Don’t plug in numbers until the end
Watch your units !
 Always check the units of your answer, and carry the units
along with your formula during the calculation.
Understand the limits !
 Many equations we use are special cases of more general
laws. Understanding how they are derived will help you
recognize their limitations (for example, constant
acceleration).
Physics 211: Lecture 1, Pg 16