9. Motion graphs (distance, velocity and acceleration against time)_

Today’s Objectives

Plot and explain the graphs of the relationship between distance, speed/velocity/acceleration and time, and calculate the area under the graph of speed/velocity in c motion:
- in constant speed/velocity;
- in constant acceleration;

Graphing !

A … Starts at home (origin) and goes forward slowly
B … Not moving (position remains constant as time progresses)
C … Turns around and goes in the other direction  quickly, passing up home

1 – D Motion

Graphing w/ Acceleration

A … Start from rest south of home; increase speed gradually
B … Pass home; gradually slow to a stop (still moving north)
C … Turn around; gradually speed back up again heading south
D … Continue heading south; gradually slow to a stop near the  starting point

A

B

C

D

Tangent Lines

x

On a position vs. time graph:

Increasing & Decreasing

Increasing

Decreasing

On a position vs. time graph:
Increasing means moving forward (positive direction).
Decreasing means moving backwards (negative direction).

Concavity

On a position vs. time graph:
Concave up means positive acceleration.
Concave down means negative acceleration.

Special Points

P

Q

R

S

Curve Summary

A

B

C

D

All 3 Graphs

Graphing Animation Link

This website will allow you to set the initial velocity and acceleration of a car. As the car moves, all three graphs are generated.

Car Animation

Graphing Tips

Line up the graphs vertically.
Draw vertical dashed lines at special points except intercepts.
Map the slopes of the position graph onto the velocity graph.
A red peak or valley means a blue time intercept.

Graphing Tips

The same rules apply in making an acceleration graph from a velocity graph. Just graph the slopes! Note: a positive constant slope in blue means a positive constant green segment. The steeper the blue slope, the farther the green segment is from the time axis.

Real life

Note how the v graph is pointy and the a graph skips. In real life, the blue points would be smooth curves and the green segments would be connected. In our class, however, we’ll mainly deal with constant acceleration.

Area under a velocity graph

“forward area”

“backward area”

Area above the time axis = forward (positive) displacement.
Area below the time axis = backward (negative) displacement.
Net area (above - below) = net displacement.
Total area (above + below) = total distance traveled.

Area

The areas above and below are about equal, so even though a significant distance may have been covered, the displacement is about zero, meaning the stopping point was near the starting point. The position graph shows this too.

Area units

Imagine approximating the area under the curve with very thin rectangles.
Each has area of height  width.
The height is in m/s; width is in seconds.
Therefore, area is in meters!

v (m/s)

t (s)

12 m/s

0.5 s

12

The rectangles under the time axis have negative  heights, corresponding to negative displacement.

Graphs of a ball thrown straight up

The ball is thrown from the ground, and it lands on a ledge.
The position graph is parabolic.
The ball peaks at the parabola’s vertex.
The v graph has a slope of -9.8 m/s2.
Map out the slopes!
There is more “positive area” than negative on the v graph.

Graph Practice

Try making all three graphs for the following scenario:
1. Schmedrick starts out north of home. At time zero he’s driving a cement mixer south very fast at a constant speed.
2. He accidentally runs over an innocent moose crossing the road, so he slows to a stop to check on the poor moose.
3. He pauses for a while until he determines the moose is squashed flat and deader than a doornail.
4. Fleeing the scene of the crime, Schmedrick takes off again in the same direction, speeding up quickly.
5. When his conscience gets the better of him, he slows, turns around, and returns to the crash site.

Kinematics Practice

A catcher catches a 90 mph fast ball. His glove compresses 4.5 cm. How long does it take to come to a complete stop? Be mindful of your units!

2.24 ms

Uniform Acceleration

When object starts from rest and undergoes constant acceleration:
Position is proportional to the square of time.
Position changes result in the sequence of odd numbers.
Falling bodies exhibit this type of motion (since g is constant).

t : 0 1 2 3 4

x = 1

x = 3

x = 5

( arbitrary units )

x : 0 1 4 9 16

x = 7

Spreadsheet Problem

We’re analyzing position as a function of time, initial velocity, and constant acceleration.
x, x, and the ratio depend on t, v0, and a.
x is how much position changes each second.
The ratio (1, 3, 5, 7) is the ratio of the x’s.

Make a spreadsheet like this and determine what must be true about v0 and/or a in order to get this ratio of odd numbers.
Explain your answer mathematically.

Relationships

Let’s use the kinematics equations to answer these:
1. A mango is dropped from a height h.
a. If dropped from a height of 2 h, would the impact speed double?
Would the air time double when dropped from a height of 2 h ?
A mango is thrown down at a speed v.
If thrown down at 2 v from the same height, would the impact speed double?
Would the air time double in this case?

Relationships (cont.)

A rubber chicken is launched straight up at speed v from ground level. Find each of the following if the launch speed is tripled (in terms of any constants and v).

max height
hang time
impact speed

3 v

9 v2 / 2 g

6 v / g

Today’s Objectives

Plot and explain the graphs of the relationship between distance, speed/velocity/acceleration and time, and calculate the area under the graph of speed/velocity in c motion:
- in constant speed/velocity;
- in constant acceleration;