89

t

v

.

v

v

o

.

t

Section 3.6Algebraic Results for Constant Acceleration

3.6Algebraic Results for Constant Acceleration

Although the area-under-the-curve technique can be applied to any

graph, no matter how complicated, it may be laborious to carry out, and if

fractions of rectangles must be estimated the result will only be approxi-

mate. In the special case of motion with constant acceleration, it is possible

to find a convenient shortcut which produces exact results. When the

acceleration is constant, the v-t graph is a straight line, as shown in the

figure. The area under the curve can be divided into a triangle plus a

rectangle, both of whose areas can be calculated exactly: A=bh for a rect-

angle and A=

12

12

bh for a triangle. The height of the rectangle is the initial

velocity, v

o

, and the height of the triangle is the change in velocity from

beginning to end,

.

v. The object’s

.

x is therefore given by the equation

.

x=v

o

.

t+

1

2

.

v

.

t

. This can be simplified a little by using the definition of

acceleration, a=

.

v/

.

t to eliminate

.

v, giving

.

x=v

o

.

t+

1

2

a

.

t

2

[motion with constant acceleration] .

Since this is a second-order polynomial in

.

t, the graph of

.

x versus

.

t

is a parabola, and the same is true of a graph of x versus t — the two graphs

differ only by shifting along the two axes. Although I have derived the

equation using a figure that shows a positive v

o

, positive a, and so on, it still

turns out to be true regardless of what plus and minus signs are involved.

Another useful equation can be derived if one wants to relate the change

in velocity to the distance traveled. This is useful, for instance, for finding

the distance needed by a car to come to a stop. For simplicity, we start by

deriving the equation for the special case of v

o

=0, in which the final velocity

v

f

is a synonym for

.

v. Since velocity and distance are the variables of

interest, not time, we take the equation

.

x =

12

12

a

.

t

2

and use

.

t=

.

v/a to

eliminate

.

t. This gives

.

x =

12

12

(

.

v)

2

/a, which can be rewritten as

v

f

2

=2a

.

x [motion with constant acceleration,

v

o

= 0] .

For the more general case where

v

o

.

0, we skip the tedious algebra

leading to the more general equation,

v

f

2

=v

o

2

+2a

.

x [motion with constant acceleration] .

To help get this all organized in your head, first let’s categorize the

variables as follows:

Variables that change during motion with constant acceleration:

x, v, t

Variable that doesn’t change:

a