unit 1 motion graphs lyzinskiphysics x t days 3 - 4

UNIT 1Motion Graphs

LyzinskiPhysics

x

t

Days 3 - 4

Day #3

* instantaneous velocity* d-t vs. x-t graphs

(what’s the difference???)

DefinitionInstantaneous Velocity (v) – the velocity of

an object at a precise moment in time.

v = lim(x/t)t0

Honors Only!!!

Just what is meant by “instantaneous” velocity?

t

ttt

t

To find the average velocity between two points in time, we find the slope of the line connecting these two points, thus finding the change in position (rise) over the change in time (run).

As the two points move closer together, we find the average velocity for a smaller time interval.

As the two points move EVEN CLOSER together, we find the average velocity for an EVEN SMALLER time interval.

Finally, “in the limit” that the time interval is infinitely small (or approximately zero), we find the velocity at a single moment in time.

Hence the term

“instantaneous velocity”

To find instantaneous velocities, we still use the concept of

slope. But we find the slope OF THE TANGENT TO THE

CURVE at the time in question

Definition

Tangent to a Curve – a line that intersects a given curve at exactly one point.

Good Tangents

Bad Tangents

How to find the instantaneous velocity of a specific time interval from an x-t graph …

x(m)

10 20 30 40 50

t (s)

30

20

10

0

Draw the tangent to the curve at the point in question. Then, find the slope of the tangent.

Slope = rise/run = 15 m / 9 s = 1.7 m/s (approx)

Example:

What is the instantaneous velocity at t = 20 seconds?

YOU MUST SPECIFY WHICH POINTS YOU USED WHEN FINDING THE SLOPE!!!!

(24, 30)

(15, 15)

How to find the instantaneous velocity of a specific time interval from an x-t graph …

x(m)

10 20 30 40 50

t (s)

30

20

10

0

Example:

What is the instantaneous velocity at t = 5?

If the pt lies on a segment, find the slope of the segment.

Slope = 5 m / 10 s = 0.5 m/s

(0,5)

(10,10) YOU MUST SPECIFY WHICH POINTS YOU USED WHEN FINDING THE SLOPE!!!!

How to find the instantaneous velocity of a specific time interval from an x-t graph …

x(m)

10 20 30 40 50

t (s)

30

20

10

0

Draw the tangent to the curve at the point in question. Then, find the slope of the tangent.

Slope = 0 (object at rest)

Example:

What is the instantaneous velocity at t = 25 seconds?

x-t graphs

t (sec)

x (m)

t1 t2 t3

x2

x1

x3

1

2

3

0

01

011to0 tt

xxv

Slope of line segment

02

022to0 tt

xxv

Slope of line segment

linetangent

ofslopev 3ptat

Tangent to the curve has a slope of -26m / 13.5s = -1.93 m/s THEREFORE, v = -1.93 m/s and s = 1.93 m/s

(approximately)

Open to in your Unit 1 packet1

Tangent to the curve has a slope of +22m / 22sec = 1 m/s

(13.5,-20)

(0,6) (33,2)

(11,-20)

HONORS ONLY

Given a function for position, like

x(t) = 3t2 + 3t – 6,

find the instantaneous velocity at a give time (like t = 2 sec) using your graphing calculator.

Step #1) type the equation into “y =“ as y = 3x2 + 3x - 6

Step #2) Use “2nd-Trace-6” to use the “dy/dx” function

Step #3) Since “dy/dx” is really dx/dt, type in “2” and hit enter to get the instantaneous velocity at t = 2 sec.

ANSWER: dx/dt = 15 m/s.

Day #4

* Average Acceleration

Definition• Average Acceleration ( a ) – the change in

an object’s velocity divided by the elapsed time……..IN OTHER WORDS, the rate of change of an object’s velocity.

a = v/t

Find the acceleration of each object 1) An object is moving at 40 m/s when it slows down to 20 m/s over a 10 second

interval.

2) An object is moving at -40 m/s, and 5 seconds earlier it was moving at -10 m/s.

3) An object travelling at -10 in/min is moving at +20 in/min 2 minutes later.

4) An object moving at -30 mph is moving at -20 mph 10 hours later.

a = v/t = (20 – 40m/s) / 10sec = -2 m/s2

a = v/t = [-40 – (-10m/s)] / 5sec = -6 m/s2

a = v/t = [20 – (-10 in/min)] / 2min = +15 in/min2

Slows downNegative accel

Speeds upNegative accel

Speeds up+ Accel

a = v/t = [-20 – (-30 mi/h)] / 10hr = + 1mi/h2 Slows down+ Accel

So, how can something slow down and have a positive acceleration?

An object moving at -30 mph is moving at -20 mph 10 hours later.

Its speed (30mph vs. 20 mph) clearly decreases.* remember, speed is |v|

As time marches on, the velocity become MORE positive (b/c -20mph is more positive than -30mph) THEREFORE, v is +

What do the “units” of acceleration mean?

m/s2

3 m/s2 means that your velocity increased by 3 m/s every second.

-0.1 km/min2 means that your velocity decreased by 0.1 km/min every minute that you were moving.

m/s/s m/s per second

t (sec) v (m/s)

0 0

1 3

2 6

3 9

4 12t (min) v (km/min)

0 0.3

1 0.2

2 0.1

3 0

4 -0.1

The Key Equations

12 xxx Displacement:

Velocity:

t

xx

t

xv

12

t

xvv

t

0

inst lim

Acceleration:

t

vv

t

va

12

t

vaa

t

0inst lim

AVERAGE INSTANTANEOUS

Really just tangents to the curve at a point.

Acceleration Practice Problems A car has an initial velocity of -6 m/s (at t = 0 seconds). If its acceleration is a constant +5 m/s2, find the car’s speed and velocity after each second of travel.

t = _____ seconds speed (m/s) velocity (m/s)

1

2

3

4

5

Open to in your Unit 1 packet2

-1 m/s1 m/s

+4 m/s4 m/s

+9 m/s9 m/s

+14 m/s14 m/s

+19 m/s19 m/s

A ball travelling at a speed of 5 m/s toward a wall bounces back off the wall with a speed of 4.8 m/s. If the ball was in contact with the wall for 2 ms (milliseconds), find the acceleration of the ball. A car travelling at 60 mi/h slams on its brakes. It slows down with an acceleration of magnitude 5 mi/h/s. Find the time needed for the car to come to a stop.

Call right + and left –

v1 = 5 m/s right = + 5 m/s v2 = 4.8 m/s left = -4.8 m/s

a = (v2 – v1) / t = (-4.8 – 5) / .002sec = -4,900 m/s2 = 4,900 m/s2 left

v1 = 60 mi/h v2 = 0 mi/h a = -5 mi/h2

a = (v2 – v1) / t -5 = (0 – 60) / t t = 12 s

For each motion below, draw an appropriate x-t and v-t graph

at rest const + vel const + accel speeding up const – vel slowing down const - accel