UNIT 1Motion Graphs

LyzinskiPhysics

x

t

Days 7 - 10

Day #7

* a-t graphs* “THE MAP”

x

t

UNIFORM Velocity

Speed increases as slope increases

x

tObject at REST

x

t

Object Positively Accelerating

x

t

Object Negatively Accelerating

x

t

Moving forward or backward

x-t ‘s

x

t

Changing Directionx

t

Object Speeding up

v

t

UNIFORM Positive (+) Acceleration Acceleration increases

as slope increases

v

t

UNIFORM Velocity

(no acceleration)

Object at REST

v

t

Changing Direction

v-t ‘s

v

t

UNIFORM Negative (-) Acceleration

v

t

a

t

UNIFORM Acceleration

UNIFORM Velocity

OR

An Object at REST

a-t ‘s

a

t

Graph Re-CapType of graph

Slope of a line segment

Slope of the tangent to the curve at a

point

x-t v-t a-t

Average velocity Instantaneous Velocity

Average acceleration Instantaneous acceleration

JERK!!!!!!!!!!! (no jerks on test )

Area under curve

x-t

v-t

a-t

Tells you nothingDisplacement (x)

Change in velocity (v)

a-t graphs

t (sec)

a (m/s2)

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!!!!

x-t v-t a-t

SLOPE

AREA

“3 towns, 4 roads”

x v

1) On the a-t graph, what is the instantaneous acceleration at 5 seconds?

2) On the a-t graph, what is the change in velocity during the 1st 8 seconds.

a = 6 m/s2

1)

2)

v = area = 4 (6) + (2.5 * -9) = 1.5 m/s

Open to in your Unit 1 packet5

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.

2) On the a-t graph, name each different motion interval (hint: there are 4 different answers)

Either at rest or at a constant velocity, const + accel, const – accel, non-constant – accel, non-constant + accel

3)

4)

v = area = 2.5 (6) = 15 m/s v = v2 – v1 v1 = (-5) – 15 = -20 m/s

Open to in your Unit 1 packet5

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

t

x

t

Take out your Green HandoutGiven the x-t graph below, sketch the missing v-t graph. You may assume that all accelerations are constant.

x (in)

t (s)

t (s)

v (in/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)

t (h)

v (mi/h)

Given the information below, sketch each of the missing motion graphs. Assume that v1 = 2 m/s

t (min)

v (ft/min)

t (min)

a (ft/min/min)

Figures for In-Class Notes

Use the v-t graph from the previous page to construct the appropriate a-t graph a (mi/h/h)

t (h)

t (h)

v (mi/h)

Drawing an x-t from a v-t

t (hr)

8

4

0

-4

-8

v (km/hr)

t (hr) 4 8 12 16 20 24 28

80

60

40

20

0

x (km)

Find the area under the curve in each interval to get the displacement in each interval

4 8 12 16 20 24 28

Use these displacements (making sure to start at xi, which should be given) to find the pts on the d-t curve

4m 24m 40m 18m 3m

0m - 18m -10m

“Connect the dots” and then CHECK IT!!!!

Drawing a v-t from an x-tFind the slope of each “Non-curved” intervalx (yd)

10 20 30 40 50 t (min)

40

30

20

10

0

-10

-20v (yd/min)

10 20 30 40 50 t (min)

6 yd/min -4 yd/min

0 yd/min

Plot these slopes (which are average

velocities)

For curved d-t regions, draw a sloped segment

on the v-t

+ accel. region (+ slope)6

4

2

0

-2

-4

-6

This only works if the accelerations on the x-t graph are assumed to be constant