To answer this question one must be familiar with the
equations of one dimensional motion for an object undergoing uniform
acceleration.
Acceleration is defined as the rate of change
in an object's velocity. That is, acceleration occurs when an object's velocity changes
and the amount of acceleration is proportional to the change in velocity and inversely
proportional to the time it takes to cause the change. In one dimensional motion, the
acceleration will result in either an increase in speed or a decrease in speed (commonly
called "deceleration").
The equation for acceleration in
one-D motion is
a = (Vf - Vi)/T where Vf is the final
speed, Vi is the initial speed, and T is the time it takes to cause the
change.
From this we can calculate the acceleration of the
car required to cause it to stop:
a =(0 m/s -
15.6m/s)/2.30s = -6.7826 m/s^2
With the acceleration it is
now possible to calculate the distance it will take the car to stop. There is more than
one equation that will work:
D = ViT +
(1/2)aT^2
or
Vf^2 = Vi^2 +
2aD
The first equation requires less
algebra.
D = (0m/s)X2.30s +
(1/2)(-6.7826m/s^2)(2.30s)^2
D = -17.94
m
At this point one should round the answer to the proper
number of significant digits. Rounding should be done as the last step before stating
the final answer. From the given information, we see that each measured value had only
3 significant digits. Therefore the answer needs to be rounded to three
digits.
D = -17.9 m or one could say the driver must apply
her brakes 17.9 meters before the stop sign in order to get the car to stop in
time.
If one chooses to use the second equation to
determine the distance it is first necessary to do algebra an isolate the variable
D.
Vf^2 = Vi^2 + 2aD recalling that Vf =
0m/s
0 m/s =Vi^2^ +2aD moving Vi^2 to the other side of
the equality
2aD = -Vi^2 dividing both sides by
2a
D = -Vi^2/2a
D =
-(15.6m/s)^2/(2X-6.7826)
D = 17.94 m = 17.9 m ahead of the
stop sign (as before).
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