APPENDIX 1: Distance versus Time under Constant Acceleration:
Equivalence
of Potential and Kinetic Energy
Velocity v is the rate at which
distance is changing; that is, it is the time derivative of distance:
v = dx/dt
(1)
The variable v is written in boldface type to indicate
that it is a vector: it has both magnitude and direction. Distance x
is a vector pointing from the origin to the present location of the particle or object;
it is called the "position vector." Time (t) is a scalar; it has magnitude but no
direction and is therefore written in plain type. Acceleration (a)
is the time rate of change of velocity. Thus, it is the time derivative of the velocity
vector, or the second derivative of distance with respect to time.
a = dv/dt
= (d/dt)(dx/dt) = d2
x/dt2
(2)
If an object starts with zero velocity (v = 0) at
time zero (t = 0) and then accelerates with constant acceleration (a),
its velocity at time t will be simply v = at.
To calculate the distance traveled by the object between time t = 0 and time t,
we must divide the time interval 0 to t into a series of very small intervals, each
of time length dt. The distance traveled during the interval dt is simply the velocity
at that time multiplied by the time interval:
dx = vdt
= (at)dt (3)
In the second step, we have substituted the constant acceleration relationship v
= at from above. Now to compute the total distance
traveled, we must sum the distance from all of the small dt time intervals that occur
between 0 and t.
x = σ(at)dt
(4)
If we allow the length of the time interval dt to approach zero, the summation process
becomes the integral with respect to time from time
zero to time t.
Thus, an object starting from rest at t = 0 and moving with constant acceleration
(a) will move a distance ½ at2
in time t. If this is a falling object (in a vacuum, where there is no air resistance),
a = g = 9.8 m/sec2
, and the distance formula becomes
x = ½gt2
= (4.9)t2
(6)
In the first second the object falls 4.9 m; at the end of 2 seconds, it has fallen
19.6 m; after 3 seconds, 44.1 m; and so on. The kinetic energy (KE) of a moving
object is
KE = ½mv2
(7)
where m is the mass of the object and v is the magnitude of its velocity (also called
speed). Note that v used in this sense is in plain
type, not boldface. Consider again the falling object that started at position x
= 0 at time t = 0 and falls a distance h:
Because velocity v = at = gt, we have by substitution from Equation 8
The kinetic energy is given by
KE = ½mv2
= ½m(2gh) =
mgh (10)
Now let us consider the work required to lift the
fallen object back to its original height of h. Recall that work is defined as the
force exerted times the distance over which the force acts: W = Fd. The force of
gravity acting on our object is Fg
= mg, so the work required to lift
it the distance (h) back to x = 0 is
W = Fd = (mg)d = mgh (11)
Thus, the work required to restore the object equals the kinetic energy possessed
by the object at the bottom of its fall, KE = W = mgh. When we lift the object from
x = h back to x = 0, we have increased its potential energy
(PE) by the amount mgh. This potential energy can be converted to kinetic energy
by allowing the object to fall the distance h. This type of potential energy, called
gravitational potential, has a value that clearly
depends on where we locate the origin of our coordinates, x = 0. However, it is
the change in potential energy, not its absolute
value, that the potential energy/kinetic energy balances:
ΔKE = −ΔPE (12)
The change in kinetic energy is equal to and of opposite sign to the change in potential
energy.
