Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2006

ft

min

minutes

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?

200ft

200ft

200ft

100ft

200 200 200 100 700 700 feet

ft

min

minutes

What is the displacement of the bee?

200ft

-200ft

200ft

-100ft

200 200 200 100 100

100 feet towards the hive

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.

Displacementb

aV t dt

Distance Traveledb

aV t dt

To find distance traveled we have to use absolute value.

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.

velocity graph

position graph

1

2

1

2

1

2

Displacement:

1 11 2 1

2 2

Distance Traveled:

1 11 2 4

2 2

Every AP exam I have seen has had at least one problem requiring students to interpret velocity and position graphs.

Linear Motion

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.

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

c) Find the total distance traveled by the particle.

2

16V(t) = 2t - , 1 t 4

t

Linear Motion

4 2

1

16Total distance = x + dx = 13

x

4

21

16S(4) = 2t - dx = 3 + 8 = 11

t

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.

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

c) Find the total distance traveled by the particle.

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

2

16V(t) = 2t - , 1 t 4

t

Effects of AccelerationA 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?

b) How far did the car travel during those 8 seconds?

Effects of Acceleration

8 2 800

Velocity = 5 + 2.4 t dt = 5 + 1.2 t ] = 81.8 mph

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?

b) How far did the car travel during those 8 seconds?

8 8 2

0 0

83

0

v(t) dt = 5 + 1.2t dt

= 5t + .4t

= 244.8 mph/(seconds per hour)

1 = 244.8 = .068 mile

3600

In the linear motion equation:

dSV t

dt V(t) is a function of time.

For a very small change in time, V(t) can be considered a constant. dS V t dt

S V t t We add up all the small changes in S to get the total distance.

1 2 3S V t V t V t

1 2 3S V V V t

S V t t We add up all the small changes in S to get the total distance.

1 2 3S V t V t V t

1 2 3S V V V t

1

k

nn

S V t

1n

n

S V t

S V t dt

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 1.1tC t

where t is the number of years since 1970 and C is in millions of bushels per year.

For a small , the rate of consumption is constant.t

The amount consumed during that short time is . C t t

Example 5: National Potato Consumption

2.2 1.1tC t

The amount consumed during that short time is . C t t

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

total consumption C t dt

4

22.2 1.1tdt

4

2

12.2 1.1

ln1.1tt

From the beginning of 1972 to the end of 1973:

7.066 million bushels

Work:

work force distance

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

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

k = spring constant

Hooke’s law for springs: F kx

Example 7:

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

F=10 N

x=2 M

10 2k

5 k 5F x

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

F(x)

x=4 M

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 F x dx

5 dw x dx

5 dw x dx 4

05 W x dx

42

0

5

2W x

40W newton-meters

40W joules

5F x x

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?

A Bit of WorkIt 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?

99 2

0 0

F 4 = 16

= 4k

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

Work done = 4x dx = 2x = 162 N m