unit 1 motion graphs

UNIT 1Motion Graphs

LyzinskiPhysics

x

t

Days 5 - 6

Day #5

* Acceleration on x-t graphs* v-t graphs* using v-t graphs to get a

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

x

t

x

t

x

t

x

t

POSITIVE OR NEGATIVE ACCEL? SPEEDING UP OR SLOWING DOWN?

• Slope of the tangent gives vinst

• Getting more sloped speeding up

• Getting more + sloped + Accel

• Getting less sloped slowing down• Slopes are getting less + - Accel

• Getting less sloped slowing down

• Slopes are getting less – + Accel

• Getting more sloped speeding up

• Slopes are getting more – - Accel

An easy way to remember it

I’m Negative!!!I’m Positive!!!

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

v(m/s)

2 4 6 8 10 12 t (s)

8

6

4

2

0

-2

-4

v-t graphs

Constant + accel (speeding up)

Constant + Vel (constant speed)

Constant negative accel (slowing down)

At rest

Constant negative accel (speeding up)

Constant + accel (slowing down)

Constant - Vel

v(m/s)

2 4 6 8 10 12 t (s)

8

6

4

2

0

-2

-4

How to get the velocity (v) at a certain time off a v-t graph

Example:

What is the velocity at t = 8 seconds?

Go over to t = 8.

Find the pt on the graph.

Find the v value for this time. -2 m/s

v(m/s)

2 4 6 8 10 12

8

6

4

2

0

-2

-4

Finding the average acceleration on a v-t graph

A2-4 = (v2 – v1) / t

= rise / run

= 0 m/s2 A4-10 = (v2 – v1) / t

= rise / run

= -7 / 6 = -1.17 m/s2

Example:

What is the average acceleration between 0 & 2, 2 & 4, and 4 & 10 seconds?

a0-2 = (v2 – v1) / t

= rise / run

= +4/2 = +2 m/s2

v-t graphs

t (sec)

v (m/s)

Slope of any segment is the AVERAGE acceleration

The slope of the tangent to the curve at any point is the INSTANTANEOUS acceleration

t0 t1

1) On the v-t graph, find the instantaneous velocity and instantaneous speed at 24 seconds.

2) On the v-t graph, what is the average acceleration from … a) 0 to 16 sec

b) 16-48 sec

30

20

10

0

-10

-20

-30

a = slope = (+57 m/s) / 32sec = +1.78 m/s2 (approx)

Open to in your Unit 1 packet3

v = -30 m/s s = 30 m/s

a = slope = (-30 m/s) / 16sec = -1.875 m/s2

1) On the v-t graph, at what position does the object start out (what is its initial position)? ______

2) On the v-t graph, name all the times (or time intervals) during which the object is at rest.

3) On the v-t graph, name each different motion interval.

30

20

10

0

-10

-20

-30


You can’t say. You know its speed at the start, but not where it is

Const – accel (object speeds up), const – vel, const + accel (slows down), const + accel (speeds up), const – accel (slows down)

Object is at rest whenever it crosses the t-axis t = 0, 36, 80 sec

3)

4)

5)

Day #6

* v-t graphs* slopes & areas of v-t graphs* instantaneous accelerations

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 Quick Review• The slope between 2 points on an x-t graph gets you the

_______________.

• The slope at a single point (the slope of the tangent to the curve) on an x-t graph gets you the ____________.

• The slope between 2 points on a v-t graph gets you the ____________.

• The slope at a single point (the slope of the tangent to the curve) on a v-t graph gets you the ____________.

Average velocity

Inst. velocity

Inst. accel.

Avg. accel.

NEW CONCEPTWhen you find the area “under the curve” on a v-t graph, this gets you the displacement during the given time interval.

This is NOT the area under the curve

The “area under the curve” is really the area between the graph and the t-axis.

v

t

v

t

Find the area under the curve from ….

a)0-4 seconds.

b)4-6

c)6-10

d)0-10

A = ½ (4)(-10) = -20m

v

t

4 6 10

15

-10

A = ½ (2)(-10) = -10m

A = ½ (4)(15) = 30m

A = -20 + (-10) + 30 = 0m

The displacement during the first 4 seconds is -20m

The displacement during the next 2 seconds is -10m

The displacement during the next 4 seconds is 30m

The OVERALL displacement from 0 to 10 seconds is zero (its back to where it started)

v(m/s)

2 4 6 8 10 12t (s)

8

6

4

2

0

-2

-4

How to find the displacement from one time to another from a v-t graph

Example:

What is the displacement from t = 2 to t = 10?

Add the positive and negative areas together

x = 16 m + (-6.75 m) = 9.25 m

Find the positive area bounded by the “curve”

12 m + 4 m = 16 m

Find the negative area bounded by the “curve”

(-2.25 m) + (-4.5 m) = - 6.75 m

v(m/s)

2 4 6 8 10 12t (s)

8

6

4

2

0

-2

-4

How to find the distance traveled from one time to another from a v-t graph

Example:

What is the distance traveled from t = 2 to t = 10?

Add the MAGNITUDES of these two areas together

distance = 16 m + 6.75 m = 22.75 m

Find the positive area bounded by the “curve”

12 m + 4 m = 16 m

Find the negative area bounded by the “curve”

(-2.25 m) + (-4.5 m) = - 6.75 m

v(m/s)

2 4 6 8 10 12t (s)

8

6

4

2

0

-2

-4

How to find the average velocity during a time interval on a v-t graph

Example:

What is the average velocity from t = 2 to t = 10?

The AVG velocity = x / t = 9.25 m / 8 s = 1.22 m/s

The DISPLACEMENT is simply the area “under” the curve.

x = 16 m + (-6.75 m)

= 9.25 m

12 m + 4 m = 16 m

(-2.25 m) + (-4.5 m) = - 6.75

v(m/s)

2 4 6 8 10 12t (s)

8

6

4

2

0

-2

-4

How to find the final position of an object using a v-t graph (and being given the initial position)

Example:

What is the final position after t = 10 seconds if xi = 40 m?

x = x2 – x1 x2 = x + x1 = 13.25 m + 40 m = 53.25 m

The DISPLACEMENT during the 1st 10 sec is simply the area “under” the curve.

x = 20 m + (-6.75 m)

= 13.25 m

4 m 12 m + 4 m = 20 m

(-2.25 m) + (-4.5 m) = - 6.75

v-t graphs

t (sec)

v (m/s)

Slope of any segment is the AVERAGE acceleration

The slope of the tangent to the curve at any point is the INSTANTANEOUS acceleration

t0 t1

The area under the curve between any two times is the CHANGE in position (the displacement) during that time period.

1) On the v-t graph, find the object’s average speed during the 1st 16 seconds.

2) On the v-t graph, what is the object’s average velocity during the 1st 24 seconds?


Displacement = |area| = ½ (16)(-30) + 8 (-30) = -480mv = x/t = -480 m / 24 sec = -20 m/s

6)

7)

Distance travelled = |area| = | ½ (16)(-30) | = 240ms = d/t = 240 m / 16 sec = 15 m/s

30

20

10

0

-10

-20

-30

1) If the object’s initial position was -16 meters, what would its new position be after 44 seconds (using the v-t graph)?


Area = ½ (16)(-30) + 12(-30) + ½ (8)(-30) + ½ (8)(30) = - 600 m

8)

Find all the areas “under the curve” from 0 to 44 sec

Area = x = - 600 m x = x2 – x1 -600m = x2 – (-16m) x2 = - 616m

30

20

10

0

-10

-20

-30

v (m/s)

t (sec)

10

0 3 9

Find:

1) The average acceleration during the first 3 seconds.

2) The instantaneous

velocity at t = 6 seconds.


acceleration at t = 6 seconds.

4) The displacement

over the entire trip.

Practice with v-t graphs


+3.3 m/s2

+10 m/s

0 m/s

+75 m

Find:

5) The average acceleration for the trip.

6) The total displacement for

the entire trip.

7) The total distance travelled on the trip.

8) The new position of the

object after 10 seconds if its initial position is 5 meters.

v (m/s)

t (sec)

10

0 5 10

-2 m/s2

10

-10

0 m 50 m 30 m

5

Find:

5) The time(s) at which the object is at rest.

6) Average acceleration over

the entire trip.

7) The average velocity from 14 to 34 seconds.


acceleration at 15 seconds.

9) The instantaneous acceleration at 6 seconds.

v (m/s)

t (sec)

16

-12

14 22 34

9)

10)

11)

12)

13)14 & 34 sec

+ 2 m/s2

+.35 m/s2

+ 8 m/s

approx 0.8 m/s2

unit 1 motion graphs

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