04/19/23 Dr. Sasho MacKenzie - HK 376 1

Kinematic GraphsKinematic Graphs

Graphically exploring derivatives and integrals

04/19/23 Dr. Sasho MacKenzie - HK 376 2

Slope

X

Y

(0,0)

(4,8)8

4

Slope = rise = Y = Y2 – Y1 = 8 – 0 = 8 = 2 run X X2 – X1 4 – 0 4

04/19/23 Dr. Sasho MacKenzie - HK 376 3

Velocity is the slope of Displacement

X

Y

(0,0)

(4,8)8

4

AverageVelocity = rise = D = D2 – D1 = 8 – 0 = 8 m = 2 m/s

run t t2 – t1 4 – 0 4 s

Displacement(m)

Time (s)

04/19/23 Dr. Sasho MacKenzie - HK 376 4

1. The displacement graph on the previous slide was a straight line, therefore the slope was 2 at every instant.

2. Which means the velocity at any instant is equal to the average velocity.

3. However if the graph was not straight the instantaneous velocity could not be determined from the average velocity

04/19/23 Dr. Sasho MacKenzie - HK 376 5

Instantaneous Velocity

• The average velocity over an infinitely small time period.

• Determined using Calculus• The derivative of displacement• The slope of the displacement curve

04/19/23 Dr. Sasho MacKenzie - HK 376 6

Instantaneous Acceleration

• The average acceleration over an infinitely small time period.

• Determined using Calculus• The derivative of velocity• The slope of the velocity curve

04/19/23 Dr. Sasho MacKenzie - HK 376 7

Average vs. Instantaneous

AverageVelocity = rise = D = D2 – D1 = 8 – 0 = 8 m = 2 m/s

run t t2 – t1 4 – 0 4 s

X

Y

8

4

Displacement(m)

Time (s)(0,0)

(4,8)

The average velocity does not accurately represent slope at this particular point.

04/19/23 Dr. Sasho MacKenzie - HK 376 8

Derivative

• The slope of the graph at a single point.• Slope of the line tangent to the curve.• The limiting value of D/ t as t

approaches zero.

04/19/23 Dr. Sasho MacKenzie - HK 376 9

Infinitely small time period (dt)

dt

•Tangent line•Instantaneous Velocity

Displacem

ent

Time

t1

t2

t3

04/19/23 Dr. Sasho MacKenzie - HK 376 10

Velocity from Displacement

Displacem

ent

Time

Velocity > 0

Velocity = 0

Velocity < 0

The graph below shows the vertical displacement of a golf ball starting immediately after it bounces off the floor and ending when it lands again.

04/19/23 Dr. Sasho MacKenzie - HK 376 11

Graph Sketching Differentiation

0

+

0

Displacement Velocity

Velocity Acceleration

OR

Slope

0

0

04/19/23 Dr. Sasho MacKenzie - HK 376 12

Displacement Velocity

Velocity Acceleration

OR

0

0 0

+

0

0

_

SlopeGraph

Sketching Differentiation

04/19/23 Dr. Sasho MacKenzie - HK 376 13

Going the other way: area under the curve

Area under curve = Height x Base = Y x X = 4 x 2 = 8

Y

X

4

2(0,0)

(2,4)

04/19/23 Dr. Sasho MacKenzie - HK 376 14

Displacement is the Area Under the Velocity Curve

Displacement = V x t = 4 x 2 = 8 m

Velocity (m/s)

Time (s)

Y

X

4

2(0,0)

(2,4)

04/19/23 Dr. Sasho MacKenzie - HK 376 15

What if velocity isn’t a straight line?

This would be an over estimate of the area under the curve

(2,4)4

2(0,0)

Velocity

Time

04/19/23 Dr. Sasho MacKenzie - HK 376 16

Integration

• Finding the area under a curve.• Uses infinitely small time periods.• All the areas under the infinitely

small time periods are then summed together.

• D = Vdt = Vt

04/19/23 Dr. Sasho MacKenzie - HK 376 17

Infinitely small time periods

Velocity

Time

•D = Vdt = V1t1 + V2t2 + V3t3 + ……

InstantaneousVelocity

Infinitely smalltime period

The area under the graph in these infinitely small time periods are summed together.

04/19/23 Dr. Sasho MacKenzie - HK 376 18

Graph Sketching Integration

Displacement Velocity

Velocity Acceleration

OR

0

2

0

Constant slope of 2

t t t

Area under curve increases by the same amount for each successive time period (linear increase).

04/19/23 Dr. Sasho MacKenzie - HK 376 19

Graph Sketching Integration

Displacement Velocity

Velocity Acceleration

OR

0 t t t

Area under curve increases by a greater amount for each successive time period (exponential increase).

0

Exponential curve

04/19/23 Dr. Sasho MacKenzie - HK 376 20

Indicate on the acceleration graph below the location of the following point(s). Place the letter on the graph.

A. Zero velocity B. Zero acceleration C. Max velocityD. Min velocity E. Max acceleration F. Min acceleration

0

1 s 2 s 3 s

-1

2

04/19/23 Dr. Sasho MacKenzie - HK 376 21

Indicate on the velocity graph below the location of the following point(s). Place the letter on the graph.

A. Zero velocity B. Zero acceleration C. Max velocityD. Min velocity E. Max acceleration F. Min accelerationG. Max displacement H. Min displacement

0