Understanding Motion
Linear Motion
Motion
• The motion of an object can only be recognized
when something is established as a basis of
comparison…a reference point
• We say an object is moving when its position
changes compared to that reference point
• For most day-to-day situations, the Earth, and
those things affixed to it, serve as a convenient
reference frame.
Position and Time
• These are the two most fundamental physical
quantities that can be measured to describe
an object’s motion.
• The relationship between these variables can
be discovered experimentally and modeled
using mathematics in both graphical and
equation form.
Position vs. Time Graphs
• The magnitude (size) of the slope tells us…
• The algebraic sign of the slope tells us…
• The magnitude and sign together tell us…
• The vertical intercept tells us…
• When this graph is a straight line we know…
The generic equation for a
linear graph is…
y = mx + b
In terms of the physical
quantities being plotted this
becomes…
x = mt + b
Position, x (m)
Position vs. Time Graphs
Time, t (s)
If we replace the slope and
intercept terms with what
they tell us we get…
x = vt + xo
Where v is the velocity (m/s)
and xo is the initial position
(m) of the object in motion
Position, x (m)
Position vs. Time Graphs
m=v
xi
Time, t (s)
x = vt + xo
• This equation (straight line with slope v and
intercept xo) is a model that describes the
relationship between position and time for an
object moving with constant velocity.
x = position of object after time, t
v = velocity of object (speed in a direction)
t = elapsed time
xo = starting position of the object
Ball on a ramp questions:
1. What information does the slope of an x-t graph tell
us?
2. Does your x-t graph for the ball on a ramp have slope?
3. What does the shape of your x-t graph tell us about
the motion of the ball?
4. Does the information from your v-t graph support the
description given in #3? How so?
5. What information does the slope of a v-t graph tell
us?
6. What information does the intercept of the v-t graph
tell us?
7. What value would you expect for the intercept in this
lab activity? Why?
Constantly Accelerated Motion
(ball on a ramp)
• Velocity-time graphs are linear… v  t
– Slope is constant  The rate at which velocity is
changing is constant SLOPE = ACCELERATION
• Position-time graphs are NOT linear, they are
quadratic… x  t2
– slope is NOT constant  Velocity is changing
vf = at + vo
• This equation is a model describing the
relationship between velocity and time for an
object that is constantly accelerating
vf – “final” velocity after time, t
a – acceleration (slope of v-t graph)
t – elapsed time
vo – starting velocity (intercept of v-t graph)
x=½
2
at
+ vot + xo
• This equation models the relationship
between position and time for constantly
accelerated motion
• This equation emerges from our ball on a
ramp data or it could be derived (we will!)
x = position after time, t
a = acceleration
vo = starting velocity
t = elapsed time
xo = starting position
A summary of motion equations thus
far…
Constant velocity:
x = vt + xo

x = vt
or v= x/t
(v is constant or
average velocity)
Constant acceleration:
vf = at + vo

x = ½ at2 + vot + xo

a = v/t
x = ½ at2 + vot