LINEAR MOTION AND GRAPHS

Dr. Cahit Erkal

We will study two types of graphs: displacement-time and velocity-time. We will then learn how to obtain a velocity versus time graph from the displacement-time graph (also the reverse process). This presentation can be adapted to any grade level between 5 and 12 with some modifications. It presents the most fundamental concept of physics, the linear motion with minimal mathematics, while providing a test-ground for the skills developed in mathematics classes. The goal of this exercise is to introduce physics early on, with the hope that, it will reinforce some of the mathematical concepts learned--while encouraging the teachers to include at least one unit of description of motion.

__Arithmetic skills practiced in this exercise:__

Number line, concept of direction on the number line (coordinate frame is the fancy name), slope, finding areas of such geometric shapes like triangle or rectangle, graphing, tables (charts).

Car-1

**. **** A**

A
denotes the displacement of Car-1

B
denotes the displacement of Car-2

Both cars are at the same distance from origin, however they
have different
displacements.

2) Let’s consider a coordinate frame (two meter sticks placed mutually perpendicular), which measures the displacements in horizontal (x) as well as vertical (y) directions.

4

A
denotes the position of the car at this 3 A

particular
time; B denotes the position

of the car at a later time.
2 B

1

x (m)

x,y

A: (2,3) B: (4,2) `1
2 3 4

During the time the car’s displacement changed from A to B, its x-displacement changed from 2 to 4, its y-displacement changed from 3 to 2.

__Graphs__

Slope of a straight line = Rise/Run

You can calculate the slope by choosing any

two points on the line [such as A (X_{1},Y_{1})
and

X X

B (X

_{2}-X_{1}=Run_{1 }X_{2}_{2}, Y_{2})].

_{ } X_{}

y_{2}-y_{1}

Slope =

x_{2}-x_{1}

Note that the application of slope in describing the velocity is accomplished by taking the horizontal axis as time(t) and the vertical axis(y) as displacement x.

x_{2}-x_{1}

Velocity V =
velocity is the slope of x-t graph

t_{2}-t_{1}

Let’s
record the displacements of a toy car as it moves in one dimension [it can go
the positive x-direction (to the right)
or in the negative x-direction (to the left).

__Time (t)__ __Displacement
(cm)__

0 0

5 4.1

10 7.9

15 12.1

20 16.0

25 16.0

30 16.0

35 18.0

40 20.1

45 21.9

50 24.0

55 22.1

60 20.0

70 0

Note
that the motion on x-t graph is described by a set of straight lines. We analyze the motion by breaking the whole
graph into segments, each of which describes the motion between the break point
(20 sec, 30 sec, 50 sec, 60 sec).
Realistically speaking, the lines would smoothly join to each other in
an actual experiment.

Velocity
= (x_{2}-x_{1})/(t_{2}-t_{1})

__ t _{2}-t_{1} (s)__

20-0 16-0 16/20
= 0.8

30-20=10 16-16 = 0 0/10 = 0

50-30=20 24-16 = 8 8/20 = 0.4

60-50=10 20-24 = -4 -4/10
= -0.4

Now we are ready to plot (or graph) v – t.

At the end points (20,30,50,60 seconds) velocity is
not defined! Because the lines on x-t
graph break , slope is undefined.

Let’s
plot the two graphs together. We can
better see how the change in the displacement reflects on the v-t graph.

Note:

- In each time interval
velocity is constant (not changing), therefore acceleraton is zero

- From 20 to 30 sec.
velocity is zero

- When velocity is
positive (0-20 sec., 30-50 sec.) the toy car is moving in positive
x-direction

- When velocity is
negative (50-60 sec.) the toy car is moving in negative-x-direction

__Velocity-time
graph (v-t graph):__

V-T
graph displays the variation of an object’s velocity as a function of
time. The slope of the v-t graph is
acceleration a. Acceleration is the
rate of change of velocity. It’s unit
is cm/s^{2 }or^{ }m/s^{2}_{. }We will study only constant
accelerated motion.

1)

3-2 40-20

v(cm/s)5

4

3

2

1

10
20 30 40

t(s)

slope
= _{}

This means that velocity of the
object __changes by 0.05 cm/s every second__.

2)

v (cm/s) 5 3 20 30 t (s) 10 15
-5

_{}

Let’s assume v-t graph is
given and ask the question “what is the displacement from time t_{1},
to time t_{2}?”

v(cm/s)

10

7

t(s)

10 20
30

-8

This is a negative area; height of the rectangle is –8 ,
width is 10 seconds.
It represents a negative displacement, which means that the object
is now displacing toward the origin.
Its direction of motion is opposite to its initial displacement. Area of this rectangle is the displacement from t = 0 to
t = 10s

time(s) |
Displacement
(cm) |

0-10 |
area
= 10x10 = 100 |

10-20 |
area
= -8x(20-10) = -80 |

20-30 |
area
= 7(30-20) = 70 |

Total
distance covered = 100+80+70 = 250 cm

Total
Displacement = 100 +(-80) + 70 = 90 cm, this means that object is 90 cm away
from the origin, in the positive direction.

**Reverse
Problem: Given the v-t graph, sketch the motion:**

In this part of this exercise, you can ask the students to simulate the motion by moving forward and backward along a straight line, or use a remote controlled toy car alongside a meter stick. The idea is to be able to identify the information provided on the graph with the real life situation. If students can see how motion can be analyzed using real life examples, they process the information more effectively.

We can sketch the approximate motion of the object for which the v-t graph is known. Take the example we just worked out:

0-10 s : object has initial velocity of 10 cm/s and displaces 100 cm in the first 10 sec.

10-20s: at t=10 sec it reverses (its velocity becomes negative) direction and covers 80 cm in the negative x direction which maintaining 8 cm/s velocity in the negative direction.

20-30 s: at t = 20 s it reverses direction and moves 70 cm in the positive x direction.

We can now sketch this as movements either along the horizontal number line: