Answer 2:
To answer this problem, one must first
recognize that distance traveled is the
integral of velocity, and velocity is the integral
of acceleration. Therefore,the distance
traveled can be calculated as the area under
the curve describing the relationship between
velocity and time.
Conversely, acceleration is the
derivative of velocity, and velocity is the
derivative of distance. A derivative can be
interpreted as the rate of change of a
parameter, so that velocity is the rate of
change of distance, and can be calculated as the
slope of the relationship between distance and
time.
In order to calculate the distances traveled at
times 1 and 2 seconds, the velocity must be
converted from miles/hour to miles/second using
the following conversion factors:
60 miles/hour * 1 hour/60 minutes * 1 minute/60
seconds = 1/60 miles/second
If you assume that the car has
a constant acceleration from 0 to 1/60miles/second
in 12 seconds, then you can graphically depict
this acceleration as a linear plot of velocity
versus time, where the velocity, v, at time
t = 0 seconds is:
v = 0 miles/second, and the
velocity at time t = 12 seconds is:
v =1/60 miles/second.
The relationship between velocity and time is
therefore the line
v = a*t, where v is velocity in
miles/second, t is time in seconds, and a is
the acceleration in
miles/(second)^{2}, or the slope of this
line. The acceleration is the rate of change,
or derivative, of velocity.
Since you already know two points on
this line, you can calculate the acceleration or
slope of the line. The two points (v, t) on the
line are (0 miles/second, 0 seconds) and
(1/60miles/second, 12 seconds). Therefore, 1/60
miles/second = a * 12 seconds, and acceleration,
a = 1/60 miles/second * 1/12 seconds = 1/720
miles/(second)^{2}.
The relationship between velocity and seconds
is described as
v = [1/720miles/second^{2}]*t.
As described earlier, distance is merely the
integral of velocity (since velocity is the
derivative of distance).
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