Question 1184683
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A car left skid marks 16 m long when the driver brought it to a quick stop as he was traveling east. 
Assuming the brakes decelerated the car at -15 m/s², how fast was the car traveling 
when its brakes were first applied?
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<pre>
Standard designations:

    S for the length (meters);

    t for the time (seconds);

    a for acceleration (deceleration) (m/s^2);

    v for the speed  (m/s).


Basic formulas from Physics/Mechanics/Kinematics are

    S = {{{(a*t^2)/2}}},    (1)

    a = {{{v/t}}}.          (2)


These formulas are of the first row, and it is assumed that 
every student, who learns the subject, knows them without remainder.


We are given S and "a", and they want we determine "v".


But the basic formulas (1) and (2) do not provide direct connection between S, "a" and "v"
- so, we need to derive the necessary formula from (1) and (2).


From (2), express  t = {{{v/a}}}  and substitute it into formula (1).  You will get

    S = {{{(at^2)/2}}} = {{{(a*v^2)/(2a^2)}}} = {{{v^2/(2*a)}}},

which implies

    v^2 = {{{2aS}}},  or  v = {{{sqrt(2aS)}}}.   (3)


Formula (3) is what we need.  Now, to get the answer, substitute the given data in (3)

    v = {{{sqrt(2*15*16)}}} = {{{sqrt(480)}}} = 21.91 m/s.


At this point, the problem is solved completely.


<U>ANSWER</U>.  The speed of the car was  21.91 m/s when its brakes were first applied.
</pre>

Solved.


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Formula (3) is also the basic formula of kinematics, but it is traditionally 
considered as the basic formula of the second row.