Days 7 - 10
UNIT 1
Motion Graphs
x
t
Lyzinski
Physics
Day #7
* a-t graphs
* “THE MAP”
UNIFORM Velocity
x
x
Speed increases as
slope increases
t
Object at REST
t
Moving forward or
backward
x
x
Object
Positively
Accelerating
x-t
‘s
t
t
x
x Changing Direction
x
Object
Speeding
up
t
t
t
Object Negatively
Accelerating
UNIFORM Positive (+)
Acceleration
v
Acceleration increases
as slope increases
v
t
v
Changing Direction
t
v-t
UNIFORM Velocity
‘s
t
(no acceleration)
v
v
Object at
REST
UNIFORM Negative (-)
Acceleration
t
t
UNIFORM Acceleration
a
a
t
a-t
‘s
t
UNIFORM Velocity
OR
An Object at REST
Graph Re-Cap
Type of graph
Slope of a line
segment
Slope of the tangent to
the curve at a point
x-t
Average velocity
v-t
Average acceleration Instantaneous acceleration
a-t
JERK!!!!!!!!!!! (no jerks on test )
Area under
curve
Instantaneous Velocity
x-t
v-t
Tells you nothing
a-t
Change in velocity (Dv)
Displacement (Dx)
a-t graphs
a (m/s2)
t (sec)
t0
t1
The area under the curve between any
two times is the CHANGE in VELOCITY
during that time period.
Slopes????
No jerks on test 
THE MAP!!!!
Dx
x-t
AREA
v-t
Dv
a-t
SLOPE
“3 towns, 4 roads”
Open to 5 in your Unit 1 packet
1) On the a-t graph, what is the instantaneous acceleration at 5 seconds?
1)
a = 6 m/s2
2) On the a-t graph, what is the change in velocity during the 1st 8 seconds.
Dv = area = 4 (6) + (2.5 * -9) = 1.5 m/s
Open to 5 in your Unit 1 packet
1) On the a-t graph, if the object’s final velocity after 4 seconds is -5 m/s, find its initial velocity at t = 0.
3)
Dv = area = 2.5 (6) = 15 m/s
Dv = v2 – v1
 v1 = (-5) – 15 = -20 m/s
2) On the a-t graph, name each different motion interval (hint: there are 4 different answers)
4)
Either at rest or at a constant velocity, const + accel, const – accel,
non-constant – accel, non-constant + accel
Day #8
FREE-FALL LAB
Day #9
Drawing Physics Graphs
from word-problem
scenarios
Day #10
Given x-t or v-t graphs,
draw the corresponding v-t
or x-t (or even a-t) graph.
v
x
t
t
Take out your Green Handout
Given the x-t graph below, sketch the missing v-t graph. You may assume that all accelerations
are constant.
x (in)
t (s)
v (in/s)
t (s)
Given the v-t graph below, sketch the x-t graph. You can assume that the object starts off at a
position of 0 mi.
x (mi)
t (h)
v (mi/h)
t (h)
Given the information below, sketch each of the missing motion graphs. Assume that v1 = 2 m/s
v (ft/min)
t (min)
a (ft/min/min)
t (min)
Use the v-t graph from the previous page to construct the appropriate a-t graph
a (mi/h/h)
t (h)
v (mi/h)
t (h)
Drawing an x-t from a v-t
Find the area under the curve in each interval
to get the displacement in each interval
v (km/hr)
4m
8
4
0
-4
-8
24m
40m
18m
0m
4
3m
- 18m
8
12
16
20
-10m
24
28
t (hr)
Use these displacements (making sure to start at xi, which
should be given) to find the pts on the d-t curve
x (km)
80
“Connect the
dots” and then
CHECK IT!!!!
60
40
20
0
t (hr)
4
8
12
16
20
24
28
Drawing a v-t from an x-t
Find the slope of each “Non-curved” interval
x (yd)
40
30
20
10
0
6 yd/min
10
-4 yd/min
20
30
40
50
t (min)
-10
-20
0 yd/min
Plot these slopes
(which are average
velocities)
v (yd/min)
+ accel. region
(+ slope)
6
4
2
0
-2
-4
-6
10
20
30
40
This only works if the
accelerations on the x-t
graph are assumed to be
constant
50
t (min)
For curved d-t regions,
draw a sloped segment
on the v-t