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 4 2,3

always positive

 2,  1,1,2

A honey bee makes several trips from the hive to a flower garden. The velocity graph is shown below.

What is the total distance traveled by the bee?

 700 700 feet 100 ft min 50 200ft 200ft 0 -50 -100 2 4 200ft 6 minutes 8 100ft 10 

What is the displacement of the bee?

100 feet towards the hive 100 ft min 50 200ft 0 -50 -100 2 200ft 4 -200ft 6 minutes 8 10 -100ft 

To find the displacement (position shift) from the velocity function, we just integrate the function. The negative areas below the x-axis subtract from the total displacement.

Displacement

 

a b

To find distance traveled we have to use absolute value.

Distance Traveled

 

a b

Find the roots of the velocity equation and integrate in pieces, just like when we found the area between a curve and the x-axis. (Take the absolute value of each integral.) Or you can use your calculator to integrate the absolute value of the velocity function.



2 0 -1 1 0 1 1 1 2 -1 2 1 3 2 4 2 velocity graph -2 2 1 -2 1 2 3 4 position graph 5 5 Displacement: 1 2 1 2 2 Distance Traveled: 1 1 2 1 2 2 4 Every AP exam I have seen has had at least one problem requiring students to interpret velocity and position graphs.



Linear Motion

V(t) = 2t - 16 t

2

, 1



t



4

V(t) is the velocity in m/sec of a particle moving along the x-axis and starting at the position, s(0) = 8.

a) Determine when the particle is moving to the right, to the left, and stopped.

Particle is moving left on 1 < t < 2, stopped at t = 2 and moving right on 2 < t < 4.

b) Find the particle’s displacement for the given time interval and its final position.

S(4) =  1 4 2t - 16 t 2 c) Find the total distance traveled by the particle.

dx = 3 + 8 = 11 Total distance =  1 4 x 2 + 16 x dx = 13

Effects of Acceleration

A car moving with initial velocity of 5 mph accelerates at the rate of a(t) = 2.4 t mph per second for 8 seconds.

a) How fast is the car going when the 8 seconds are up?

Velocity = 5 +  0 8 2.4 t dt = 5 + 1.2 t 2 ] 0 8 = 81.8 mph b) How far did the car travel during those 8 seconds?

 0 8 v(t) dt =  0 8  5 + 1.2t

2  dt = 5t + .4t

3  0 8 = 244.8 mph/(seconds per hour) 1 = 244.8 3600 = .068 mile

In the linear motion equation:

dS



dt V

(

t

) is a function of time.

dS

     

t

For a very small change in time,

V

(

t

) can be considered a constant.

We add up all the small changes in

S

the total distance.

to get

S

1

t V

2

t V

3

S

 

V

1

V

2

V

3 

t



  

t

We add up all the small changes in

S

the total distance.

to get

S V

1

t V

2

t V

3

S

 

V

1

V

2

V

3 

t S



n k

  1

V n

 

t S



n

   1

V n

 

t S

    As the number of subintervals becomes infinitely large (and the width becomes infinitely small), we have integration.



This same technique is used in many different real-life problems.



Example 5: National Potato Consumption The rate of potato consumption for a particular country was: 

2.2 1.1

t

where

t

is the number of years since 1970 and

C

is in millions of bushels per year.

  

Example 5: National Potato Consumption 

2.2 1.1

t

  We add up all these small amounts to get the total consumption:

total consumption

    From the beginning of 1972 to the end of 1973:  2 4

2.2 1.1

t dt



2.2

t



1 ln1.1

1.1

t

4 2 

7.066

million bushels 

Work:

work

 Calculating the work is easy when the force and distance are constant.

When the amount of force varies, we get to use calculus!



Hooke’s law for springs:

F



kx k

= spring constant

x

= distance that the spring is extended beyond its natural length 

Hooke’s law for springs:

F

=10 N

F



kx

Example 7:

x

=2 M It takes 10 Newtons to stretch a spring 2 meters beyond its natural length.

10 5 

k

2

F x

How much work is done stretching the spring to 4 meters beyond its natural length?



x

=4 M

F

(

x

) How much work is done stretching the spring to 4 meters beyond its natural length?

For a very small change in

x

, the force is constant.

dw

  5

x

dw

 

dw

 

W

 5 2

x

2 0 4

W

  0 4

W

 40 newton-meters

W

 40 joules 

A Bit of Work

It takes a force of 16 N to stretch a spring 4 m beyond its natural length. How much work is done in stretching the spring 9 m from its natural length?

  = 16 = 4k so k = 4 N/m and F(x) = 4x for this spring.

Work done =  0 9 4x dx = 2x 2 9 0 = 162 N m