Question 1183954
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A particle moving in a constant deceleration passes by two ports, which are 500 m apart, with velocity 40m/s and 20 m/s respectively. 
Find the: (a) deceleration of the particle; (b) time after passing by the second post when it comes to rest
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            It can be solved as an  Algebra problem,  OR as a  Calculus problem  OR  as a  Physics problem.

            The  Physics solution is much more elegant and much more educative,  so I will present here the  Physics solution.



<pre>
Let "m" be the mass of the particle; "a" be the constant deceleration.


Then the change of the kinetic energy between the ports is

    {{{(mv[1]^2)/2}}} - {{{(mv[2]^2)/2}}}  joules.


The deceleration force, acting on the particle, is  the product ma,  and the work of this force on the distance 
between the ports is  ma*500  joules.


So, we can write the conservation energy equation in the form

    {{{(mv[1]^2)/2}}} - {{{(mv[2]^2)/2}}} = 500ma


(change of the kinetic energy is equal to the work done).


Cancel the mass m in both parts of the equation; then substitute the given velocities. You will get

    {{{40^2/2}}} - {{{20^2/2}}} = 500a.


Simplify and find deceleration value

    {{{1600/2}}} - {{{400/2}}} = 500a

    800 - 200 = 500a  --->  600 = 500a  --->  a = 600/500 = 1.2.


Thus deceleration is  1.2 m/s^2.   It is the answer to question (a).


After passing the second port, the particle will be at rest in  20/a = 20/1.2 = 16 2/3  seconds.     It is the answer to question (b).
</pre>

Solved.