Chapter 2
Describing Motion: Kinematics
in One Dimension
• Introduction
(previous lecture)
• Reference Frames and Displacement
• Average Velocity
(previous lecture)
• Instantaneous Velocity
• Acceleration
(previous lecture)
(previous lecture)
• Motion at Constant Acceleration
• Falling Objects
(previous lecture)
Recalling Last Lecture
You need to define a reference frame in order to fully characterize the motion of an
object. In general we use Earth as reference frame for
Our measurements.
Displacement is how far an object is
from its initial position after an interval
of time ∆t:
(2.1)
x1
x2
Eq. 2.1 gives the magnitude of the vector
Average velocity is given by:
And average acceleration is:
(2.4)
(2.6)
Both velocity and acceleration are vectors (gives direction and magnitude) :
Acceleration, velocity, displacement, and time.
Deceleration
The car from the previous example, now moving to the left and
decelerating.
Few Relevant Notes
I. The magnitudes of the instantaneous velocity and instantaneous speed are the
same, though this is not necessarily true for average velocity and average speed.
II. If I give you an interval of time ∆t = t2 – t1 but do not specify the initial and final
time t1 and t2 , you can then assume t1 = 0 such that ∆t = t2 – t1 = t2 = t.
III. Similarly, if I give you a displacement ∆x = x2 – x1 but x1 is not specified, you can
assume x1 = 0 such that ∆x = x2 – x1 = x2 = x.
II and III are just a matter of redefining the origin of your coordinate system. For
Example:
0
x1
x2
X1 0
Translate the origin from “0” to “x1” by applying a transformation:
X = x – x1
such that
X1 = 0 – x1 and X2 = x2 – x1
are the new set of coordinates.
X2
Motion at Constant Acceleration
It is not unusual to have situations where the acceleration is constant. In this lecture
you will see one of such situations that you experience in your everyday life. But
before, let’s assume a motion in a straight line subject to a constant acceleration. In
this case the following is valid:
The instantaneous acceleration and the average acceleration have the same
magnitude. We can then write:
(2.8)
We will use item II from the previous slide with following conventions:
t1 = 0 ,
t2 = t
and also define the position and velocity at t1 = t = 0 as
x1 = x0 ,
v1 = v0 ,
for our future developments.
x = x2
v = v2
Motion at Constant Acceleration
With this in mind, let’s find some useful equations for situations with constant
acceleration that relate a, t, v, v0, x and x0 such that you can calculate any unknown
variable among them if you know the others.
Let’s first obtain an equation to calculate the velocity v of an object subjected to a
constant acceleration a:
Using the definitions introduced on the previous slide, we can rewrite eq. 2.6 as:
(2.9)
Eq. 2.10 can be rearranged:
(2.10)
Eq. 2.10 allows you to calculate the velocity v of an object at a given time t knowing
the its acceleration a and initial velocity v0.
Motion at Constant Acceleration
Example: A car accelerates from rest at a constant rate of 5.0 m/s2. What is its
velocity after 5.0 s?
Solution:
Known variables:
v0 = 0.0 m/s
a = 5.0 m/s2
t = 5.0 s
Using eq. 2.10:
Motion at Constant Acceleration
We can also obtain an equation that gives the position x of an object knowing its
constant acceleration a, initial position x0 and initial velocity v0:
Using our definitions, eq. 2.4 can be written as:
(2.11)
On the other hand, since the velocity of the object changes from v0 to v at a constant
rate a, we have:
(2.12)
Motion at Constant Acceleration
But (Eq. 2.11) = (Eq. 2.12). Then:
Using equation 2.10 in the above expression:
or
(2.13)
Eq. 2.13 allows you to calculate the position of an object at a given time t knowing
its acceleration a , initial velocity v0 and initial position x0 .
Motion at Constant Acceleration
There is another very useful equation that applies to motion at constant acceleration
in situations where no time information has been provided.
We will use equations 2.10 and 2.13 to eliminate the time variable:
(2.10)
(2.13)
From eq. 2.10 Using this expression in 2.13:
(2.14)
Motion at Constant Acceleration
Example (Problem 28 from the text book): Determine the stopping distances for a
car with an initial speed of 95 Km/h and human reaction time of 1.0 s, for an
acceleration (a) a = -4.0 m/s2 ; (b) a= -8.8 m/s2 ?
Solution:
We want to convert from h to s. It is also wise to convert from Km to m:
(remember to carry some extra figures
through your calculations)
Driver realizes he has to brake (v = v0)
Next, we have to define our
reference frame (coordinate system)
in order to calculate the position
where the driver starts breaking.
Driver starts braking (v = v0)
0
x0
x
cars stops (v = 0)
Motion at Constant Acceleration
Continuing…………
Note that no information on the time needed to stop the car is provided. This problem
is better approached with eq. 2.14. The position where the brakes are applied can be
obtained if we observe that the car is moving at constant velocity v0 before the driver
reacts. We can then use the equation of motion for constant velocity to find x0:
As we have already mentioned, the final velocity v is zero in both (a) and (b) cases:
v (final) = 0.0 m/s
(continue on the next slide)
Motion at Constant Acceleration
We have then the following initial conditions:
(a) The stopping distance for a = -4.0 m/s2 is:
(b) Similarly, the stopping distance for a = -8.0 m/s2 is:
Motion at Constant Acceleration
Let’s summarize the set equations we just found:
(2.10)
(2.12)
(2.13)
(2.14)
Note that these equations are only valid when the acceleration is constant.
Motion at Constant Acceleration
Example (Problem 21 from the text book): A car accelerates from 13 m/s to
25 m/s in 6.0 s. (a) What was its acceleration? (b) How far did it travel in this time?
Assume constant acceleration.
Solution:
This problem is about choosing an appropriate equation from the set of equations we
just found. For this purpose, first you have to write the parameters given in the
problem:
v0 = 13 m/s
v = 25 m/s
t
= 6.0 s
(continue on the next slides)
Motion at Constant Acceleration
(a) Here you want a knowing v0 , v and t. Which of the equations below you find more
appropriate to solve this problem?
(2.10)
(2.12)
(2.13)
(2.14)
It is clear that 2.10 is the better choice for this item.
;
Motion at Constant Acceleration
(b) Here you want the distance travelled (x – x0) knowing v0 , v , t and a (calculated
in item (a) ). Now, which of the equations below you find more appropriate to solve
this problem?
(2.10)
(2.12)
(2.13)
(2.14)
Either equation 2.12 or 2.13 is a good choice. Let’s use 2.12:
Falling Objects
We know from observation that an object will fall if we through it from a bridge. It will
in fact gain speed as it falls. In other words, it is being accelerated.
This acceleration is due to the action of the Earth gravitational force upon the object.
We will come back to gravitation in few lectures from now.
We use the letter g to denote the gravitational acceleration.
g is always different from zero at any position but the very center of the Earth.
We can say to very good approximation that g is constant:
(2.15)
Falling Objects
Some Notes:
(a) It is also clear that certain objects fall faster
than others even though we release them from
rest (v=0) at the same time and from the same
height.
The reason for this is the air resistance.
In fact, all objects experience the same
gravitational acceleration g.
(b) Note that we dealing here with constant
constant acceleration.
Therefore all equations obtained on the previous slides
apply to this case.
(c) Since the motion will be vertical, we will use the y axis instead of the x axis.
Falling Objects
Observing items (b) and (c) from the previous slide, we can rewrite the equations of
motion as:
(2.16)
(2.17)
(2.18)
(2.19)
Note: In most cases we will ignore the air resistance, unless otherwise stated.
Falling Objects
Example (Problem 35 from the textbook): Estimate (a) how long it took King Kong
to fall straight down from the top of the Empire State Building (380 m high), and (b)
his velocity just before “landing”?
Solution:
Let’s chose downwards as the positive y direction (again, this is just a convention).
Again, we have to identify the initial conditions.
380 m
g = 9.80 m/s2
+y
y - y0 = 380 m
v0 = 0.0 m/s
(continue on the next slides)
Falling Objects
(a) Here you want t knowing v0 , g and (y – y0) . Which of the equations below you
find more appropriate to solve this problem?
(2.16)
(2.17)
(2.18)
(2.19)
Equation 2.17 is more appropriated here:
Here, the minus sign is not physical since it would imply King Kong traveling back
time
Falling Objects
(b) Here you want v knowing v0 , g , (y – y0) and t. Which of the equations below you
find more appropriate to solve this problem?
(2.16)
(2.17)
(2.18)
(2.19)
Equation 2.16 is more appropriated here:
We might solve one more problem at the next class before starting chapter 3.
Assignment 1
Textbook (Giancoli, 6th edition), Chapter 2:
1) Problem 5, page 39
2) Problem 11, page 39
3) Problem 19, page 40
4) Problem 47, page 41
Please, note the Questions and Problems are
two different things in the textbook. The
assignment above includes only Problems.