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).

Displacement:

Position of an object (what you would read from the meter stick next to which a toy car is in motion on a straight line), as it moves, changes with time. This position change relative to an origin (a reference point) is called DISPLACEMENT. Displacement is a vector quantity (means it has duration as well as size), whereas distance covered is a scalar quantity (it has no direction, but has size).

Examples B

Car-1

. A

1)

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.

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

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 (X₁,Y₁) and

X₂-X₁=Run

X₁X₂

B (X₂, Y₂)].

y₂-y₁

Slope =

x₂-x₁

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.

x₂-x₁

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

t₂-t₁

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 = (x₂-x₁)/(t₂-t₁)

t₂-t₁ (s) displacement (cm) velocity (cm/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/s²orm/s²_.We will study only constant accelerated motion.

Examples

                             3-2

   40-20

            v(cm/s)5

10 20 30 40

t(s)

slope =

This means that velocity of the object changes by 0.05 cm/s every second.

v (cm/s)

5

3

                                                                   20        30 t (s)

                          10              15



-5

Area under v-t graph as displacement

Let’s assume v-t graph is given and ask the question “what is the displacement from time t₁, to time t₂?”

v(cm/s)