Chapters 2 Motion in One Dimension

Mechanics: Kinematics and Dynamics.

Kinematics deals with motion, but is not concerned with the cause of motion.
Dynamics deals with the relationship between force and motion.

Displacement

The word “displacement” implies the existence of an initial position (location) and a
final position … often of the same object but at different times.

The displacement, Δr, (as a result of an object’s motion in a period of time) is the
vector that points from the initial position toward the final position, within a suitable
frame of reference. It is the “change in position”.

The displacement of an object is not the same as the distance it travels.

Vector and Scalar

A vector is characterized by a magnitude and a direction. (velocity, displacement, ...)

A scalar quantity has only a magnitude but no direction. (mass, temperature, ...)

If the motion of an object is limited along a

straight line (its motion is one-dimensional),
the displacement of this object can have only
two directions (one being opposite to the
other). If a coordinate system is chosen for
the displacement (as in defining which
direction is positive), the displacement may
appear to be scalar-like. However, one should
note that the “sign” of the displacement
contains information on “direction”.

Δx = xf – xi
Final position Initial position
(coordinate) (coordinate)

1
 Average Velocity and Speed

average speed = distance traveled / elapsed time (scalar)

average velocity = displacement / elapsed time (vector)

Velocity and speed refer to the rate of

change of an object’s position (with
respect to some stationary reference point,
e.g. the origin).

If an object’s average velocity (/speed) is

zero for a period of time, does its average
speed (/velocity) have to be zero for that
period of time? No (/Yes)

Avg. Velocity (1D) Q Slope Of Line Through Two End Points

y 2 − y1 Δ ( y value)
slope = =
x 2 − x1 Δ ( x value)
Note: slopes may have units
average velocity

x f − xi Δx
v = =
t f − ti Δt

52.5 m − 5 m
In the figure, what is the average velocity between 1.00 s and 3.00 s? = 23.8 m / s
3 s −1 s

However, average speed = average velocity

2
 Instantaneous Velocity (1D): Slope Of Tangent

average x f − xi Δx
v = =
velocity t f − ti Δt

instantaneous velocity

Δx
v= lim Δt
Δt → 0

= slope of the tangent line to the x(t) plot

instantaneous speed = v

A straight line in the x(t) plot indicates constant velocity.

Instantaneous Velocity & Slope: Examples

What is the (instantaneous) velocity at t=2 s?

What is the instantaneous velocity at 1.0 s? (100-60) m / (1.8 s – 0.5 s) = 31 m/s

At which time is the instantaneous velocity the greatest: A, B, C or D?

Answer: A
What is the instantaneous velocity at 3.0 s? ~ 12 m/s

3
 Acceleration: the rate with which the velocity changes

Velocity is the rate of change of displacement (a vector).

Acceleration is the rate of change of velocity (also a vector).

The analysis of these two quantities is very similar.

The “rate of change” of a quantity is the “slope” of a plot of this quantity against time.

v f − vi Δv
average acceleration a = =
(vector) t f − ti Δt
Δv
instantaneous acceleration a= lim Δt
(vector) Δt → 0

Acceleration Examples

A straight line in the v(t) plot indicates constant acceleration.

A “negative acceleration” does not necessarily imply that the “speed” of an
object is reducing.

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 Equations For Constant 1-D Acceleration

“Constant acceleration” means that a is constant (fixed).

First, a trivial case …

If the acceleration is zero, the velocity does not change with time, and the
object will move with constant velocity, v.

If an object is at xo at time t=0, it will be found at position

x = xo + vt , Δx = x - xo = vt
at time t = t.

x v

t 0 t
0

Equations For Constant 1-D Acceleration

Non-zero constant acceleration: a (= constant) a

Initial velocity at t = 0 : vo

Velocity at time t is v = vo + at
0 t
The average velocity between time zero and time t is

1 1 v
v = ( vo + v ) = vo + a t
2 2 vo + at
The total displacement the object between time 0 and
time t is the product of the average velocity and the vo
time elapsed (t). If the object is assumed to be at the
origin at t=0, its position at time t is 0 t
x
1
Δx = v t = vo t + a t 2
2

0 t

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 More on Displacement in 1D

1
Δx = v t = vo t + a t 2
2

at
at2 / 2

vot vo

area of a trapezoid is
(1/2)*(top + base) * height area = Δx = vo t + at2 / 2
= (1/2) * (vo + vo + at) * t Displacement is the area under
= vo t + at2 / 2 the v(t) curve.

Equations For Constant 1-D Acceleration

v = vo + a t No x

1
Δx = ( vo + v ) t No a
2
1
Δx = vo t + a t 2 No v
2
Add one more v 2 = vo2 + 2 a Δx No t
equation

To get the position of the object at time t, we need to add the

displacement to the initial position of the object

x = xo + Δx

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 Which Equation to Use?

v = vo + a t
A car is accelerating with an initial velocity of 1 m/s and
1
a constant accleration of 5 m/s2. What is its velocity Δx = ( vo + v ) t
when it travels 50 m? 2
1
In 3 seconds, an object has accelerated from 0 to 20m/s Δx = vo t + a t 2
with a constant acceleration. How much distance has it 2
traveled during this time.
v 2 = vo2 + 2 a Δx

Vertical Motion with Constant Gravitational Accel.

Acceleration due to gravity

v = vo + (− g ) t
g = 9.8m/s2 downward
1
Or, a= - g = - 9.8m/s2 (because “up” is usually Δy = ( vo + v ) t
defined as the positive y direction.) 2
This acceleration is seen for all objects on earth, big or
small, heavy or light, in the absence of air resistance. 1
Δy = vo t + ( − g ) t 2
How to work on problems such as 2
“how high does it go?” or
v 2 = vo2 + 2(− g ) Δy
Velocity vanishes at the highest point.
“how long is it in the air?”
Displacement as specified in the question
is reached!

Equation At2 + Bt + C = 0 −B ± B2 − 4 AC Can usually tell which

t= sign (+ or - ) to use.
has solutions: 2A

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 Example Problems
1. A stone is thrown from the top of a building with an
initial velocity of 20.0 m/s straight upward, at an initial
height of 50.0 m above the ground. Determine (a) the
time needed for the stone to reach its maximum hight, (b)
the maximum height, (c) the time needed for the stone to
reach the groud.

2. In the Daytona 500 auto race, a Ford Thunderbird and

a Mercedes Benz are moving side by side down a
straightaway at 71.5 m/s. The driver of the Thunderbird
realizes that she must make a pit stop, and she smoothly
slows to a stop over a distance of 250 m. She spends 5.00
s in the pit and then accelerates out, reaching her previous
speed of 71.5 m/s after a distance of 350 m. At this point,
how far has the Thunderbird fallen behind the Mercedes
Benz, which has continued at a constant speed?

Review of Chapter 2

Displacement involves direction and distance (vector).

Average velocity = displacement/elapsed time
Instantaneous velocity = (average) velocity for an infinitesimally
small time period.
Average acceleration = velocity variation / elapsed time
Instantaneous acceleration = velocity variation / infinitesimally
small elapsed time
Equations for object undergoing constant acceleration in 1-D.
Free falling involves a constant acceleration of 9.8m/s2
downward for all objects.