There are three main ways of getting information form a graph

from the x axis and y axis

from the Slope

From the area under the curve

Looking at the first option

from the x and y axis

What ever type of graph you have will be the type of information you can get.

v vs. t from velocity graph you get information about velocity

x vx. t from position graph you get information about position

a vs. t from acceleration graph....

this can go on and on (forces, pressure, temperature, etc)

at t = 20 s x = 40 m

at t = 35 s x = -20 m

at t = 6 s velocity v = 8 m/s

Looking at the second option

slope

The information gathered from slope also is dependent on the type of graph

Recall that the slope is the 'rise over run' or change in rise divided by change in run

On an x vs. t graph this would equate to a change in position divided by change in time. This is what we call velocity.

On an v vs. t graph this would be change in velocity divided by a change in time. This is what we call acceleration.

slope at t = 5 seconds is 60 / 10, this gives a velocity of 6 m/s

slope at t = 30 s is -60 / 15, this gives a velocity of -4 m/s

velocity at 50 s is 2.67 m/s

velocity at 12.5 s is 0 m/s

slope at t = 1 s is 8/2, this gives an acceleration of 4 m/s

2

acceleration at 4 s is 0 m/s

2

acceleration at 11 s is -2 m/s

2

Third option

area under graph

Likewise the information from area comes from the type of graph

Recall an area is a two dimensional object (base x height) and the units will multiply as well

from a v vs. t graph the area is the made from the y and x axis terms (velocity * time , m/s * s) This will give you units of meters only, thus change in position. From a velocity graph the area is displacement.

from an a vs. t graph the area is change in velocity

The area up to the time of 8 seconds

6 s * 8 m/s = 48 m

.5 * 2 * 8 = 8 m

total = 56 m

If the problem told you the initial position was -10 m, then the position at 8 seconds would be 46 m

8

48

With the same initial condition, what is the location at a time of 12 seconds?

8 + 64 + 8 + 4 = 84, then - 10 = 74 m

From an acceleration graph, the area gives a change in velocity.

The area up to time 2 seconds

2 s * 1 m/s = 2 m/s

2

If the initial velocity was 3 m/s then the velocity at t = 2 s would be 5 m/s

A note on area under a graph the area is measured to the zero point on the y axis, thus this last area was only 2 m/s not 6 m/s (from the -2 up to +1)

Also there is a negative area, as shown in this example

The area up to 4 seconds

from 1 - 3 s

3 s * 1 m/s = 3 m/s

2

from 3 - 4 s

1 s * -2 m/s = -2 m/s

2

totals to

1 m/s

If initial velocity was still 3 m/s then velocity at t = 4 s would be 4 m/s

**sum up**

From position graph

Axis give location at time

Slope give velocity at time

From velocity graph

Axis give velocity at time

Slope give acceleration at time

Area gives change in position

From Acceleration graph

Axis gives acceleration at time

Area gives change in velocity

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