ACCELERATED MOTION

EQ: HOW CAN THE MOTION OF AN OBJECT BE MEASURED, DESCRIBED, ANALYZED

PG. 7

• “IT GOES FROM 0 MPH TO 60 MPH IN 3.6 SECONDS”

• ACCELERATION IS THE RATE OF CHANGE OF VELOCITY PER UNIT OF TIME

• VECTOR QUANTITY (DIRECTION MATTERS)

• CONSTANT VELOCITY (NOT CHANGING)NO ACCELERATION SO A=0

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tt

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HOW FAST CAN AN OBJECT CHANGE ITS VELOCITY?

EXAMPLE OF ACCELERATION

+

vf = +8 m/s v0 = +2 m/s

t = 3 s

2/23

28sm

t

vv

t

va

if

The wind changes the speed of a boat from 2 m/s to 8 m/s in 3 s.

EXAMPLE (NO CHANGE IN DIRECTION): A CAR CHANGES VELOCITY FROM 8 M/S TO 20 M/S IN 4 S. WHAT IS AVERAGE ACCELERATION?

Step 1. Draw a rough sketch. Step 2. Choose a positive direction (right).Step 3. Label given info with + and - signs.

+

v1 = +8 m/s

t = 4 s

v2 = +20 m/s

EXAMPLE (CONTINUED): WHAT IS AVERAGE ACCELERATION OF CAR?

Step 4. Recall definition of average acceleration.

+

v1 = +8 m/s

t = 4 s

v2 = +20 m/s

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t

vva if

4

/12

4

820

2/3 sma to the right

EXAMPLE (CHANGE IN DIRECTION) : A WAGON MOVING EAST AT 20 M/S ENCOUNTERS A VERY STRONG HEAD-WIND, CAUSING IT TO CHANGE DIRECTIONS. AFTER 5 S, IT IS TRAVELING WEST AT 5 M/S. WHAT IS THE AVERAGE ACCELERATION? (BE CAREFUL OF SIGNS.)

Step 1. Draw a rough sketch.

+

Step 2. Choose the eastward direction as positive.

vi = +20 m/s vf = -5 m/s

Step 3. Label given info with + and - signs.

E

Example 4: (Continued)

a = - 5 m/s2

a = - 5 m/s2

Acceleration is directed left, west.

+

vo = +20 m/s vf = -5 m/s

Dv = (-5 m/s) - (+20 m/s) = -25 m/s

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5

/25

ACCELERATION PROBLEM SOLVING FOR VELOCITY FINAL

In 1977 off the coast of Australia, the fastest speed by a vessel on the water was achieved. If this vessel were to undergo an average acceleration of 1.80 m/s2, it would go from rest to its top speed in 85.6 s. What was thespeed of the vessel?