One Dimensional Kinematics
Displacement (ŝ) - distance covered in a particular
direction
Velocity-- timed rate of change in displacement
 Avg. Vel = total displacement / total time
v = ∆x
∆t
Instantaneous Velocity: Velocity at any given
instant! If the object is not accelerating then the
avg. vel. = instant. vel.
3.2 A jogger is moving at a constant velocity of
+3.0 m/s directly towards a traffic light that is 100
meters away. If the traffic light is at the origin, x =
0 m, what is her position after running 20
seconds?
Homework: 2.3.3,4
2.4.6,7
Acceleration-- timed rate of change in velocity
 Avg. Acceleration = total change in
velocity / total time
a = ∆v
∆t
Instantaneous Acceleration:
the acceleration at any
instant in time
*Notice Velocity and
Acceleration vectors
aren’t necessarily in the
same direction!
11.2 A baseball is moving at a speed of 40.0
m/s toward a baseball player, who swings his
bat at it. The ball stays in contact with the bat
for 5.00×10−4 seconds, then moves in
essentially the opposite direction at a speed of
45.0 m/s. What is the magnitude of the ball's
average acceleration over the time of contact?
(These figures are good estimates for a
professional baseball pitcher and batter.)
Homework: 2.11.6,7
Graphic Representation of Motion
No motion (at rest):
a
v
x
t
t
t
Motion at constant (non-zero) velocity:
v
a
x
t
t
t
Motion with constant (non-zero) acceleration:
a
v
x
t
t
t
v
x
a
t
t
Decelerating!
t
negative
slope!
Negative
Acceleration!
v
x
t
a
t
t
Object is moving with a
given positive velocity,
slows to rest at a constant
rate and continues to
accelerate opposite to
original direction!
v
x
t
a
t
t
assume motion begins at x0 = 0
v
x
t
t
a
t
HW: Chapter 2:
C.10, 11, 12,14 ,16
6.1, 2
7.1
8.1
9.1
9.2
Instantaneous Velocity and Acceleration
A mathematical formula [ v(t) ] can often be used
for determining the velocity of an object at any
given point during its motion.
• if the object is not moving, then v(t) = 0
• if the object is not accelerating (moving with
constant velocity), then v(t) = avg. v = k
• if the object is accelerating, then the
instantaneous velocity equation can be found
through the Limiting Process
From now on: velocity mean instantaneous vel.!
The Limiting Process-- Derivatives
v = lim
∆x
∆t 0
∆t
v = dx
dt
To find a derivative of a given equation:
if x(t) = xn
if x(t) = A (constant)
then dx = nxn-1
then dx / dt = 0
dt
the derivative of a constant term is 0 !
The position of an object is given by the equation
x = 5t - 2t2 + 4t3 where x is in meters and t is in
seconds. A) Find the displacement (not the same
as how far it traveled) of the object at t = 1.0 s. B)
What the is velocity of the object when t = 2.3 s?
C) What is the acceleration when t = 1.0 s?
A) x = 5t - 2t2 + 4t3 when t = 1.0s
x = 5(1.0) - 2(1.0)2 +4(1.0)3 = 7.0 m
To find velocity, take the first derivative of the
displacement equation. Acceleration is the
derivative of the velocity equation.
B) v = dx / dt = 5 - 4t + 12t2 when t = 2.3 s
v2.3 = 5 - 4(2.3) + 12(2.3)2 = 59.3 m/s
C) a = dv / dt = dx / dt = - 4 + 24t at t = 2.0 s
a2.0 = - 4 + 24(2.0) = 44 m/s2
This object does NOT have a constant
acceleration rate. The acceleration varies with
time!
When the acceleration is constant, simple Algebra
can provide equations for x, v, a!
HW: Ch 2.13: 7, 9, 11, 14, 16
Equations for Motion with Constant Acceleration:
Variables
Equation
x vo v a t
v = vo + at
√
√
√
√
x = xo +vot +.5at2
√
√
√
√
v2 = vo2 + 2a(x - xo)
√
√
√
√
x = xo + .5(vo + v)t
√ √
x = xo + vt - .5at2
√
√
√
√ √
√
Freefall
Freefall: only gravity affects the motion of the
object.
The acceleration due to gravity varies from place
to place on the earth, but for now we shall
consider it to be a constant magnitude of:
g = 9.8 m/s2
When using this value in the previous equations,
we will adopt the following conventions:
 gravity will work along the y axis and up
will be positive
 convention dictates that up is + and down is
neg. so g is often used as -g in equations
An object is held out over a cliff that is 50.0 m
high. The object is then thrown straight up and
allowed to fall back down to the base of the cliff,
where it hits with a speed of 50.0 m/s. What
initial speed did was the object given and to what
height above the base of the cliff did it rise?
yo = 50.0 m (this means we have chosen the base of
the cliff as our origin)
y=0
vo2 = v2 - 2a∆y
v = -50.0 m/s
vo = 40.0 m/s
a = -g
vo = ?
From the highest point:
y=?
yo = 0
y - yo = (v2 - vo2) / 2a
vo = 0
v = - 50.0 m/s
a=-g
y = - 128 m
the negative sign means that the
object fell down 128 m from its
highest point!