Answer 1:
Any mechanics text book will tell you more about
your question, but I will give you some guidelines
which will help you with the calculations.. The
change of velocity (dv), with time (dt), gives you
the acceleration (a), of the car. So
a=dv/dt You also need to check for the
units: 60mph = (1/60)mile/sec, so the car goes
from 0 to 1/60mile/sec in 12sec. Therefore,
a=1/12sec*1/60mile/sec=1/720(mile/sec^2). The
distance (x), after time (t) is
x=v0t+0.5at*t. v0 is the initial velocity,
which is 0 in this case. So
x=0.5*1/720=(1/1440)miles after 1 sec, And
x=0.5*4/720=(1/360)miles after 2 sec. And so on.

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 slopeof 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 =
0seconds 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/second2]*t. As described earlier,
distance is merely the integral of velocity (since
velocity is the derivative of distance). If you
do some basic calculus and take the
antiderivative, or integral, of this equation,
then you find that d = *(1/720
miles/second)*t2 (so that d =
v).
Therefore, at 1 second, the distance
traveled is *(1/720 miles/second)*(1second) 2
= 1/1440 miles. At 2 seconds, the distance
traveled is *(1/720miles/second)*(2 second) 2
= 4/1440 miles = 1/360 miles.
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