Question 1186556
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A particle with a uniform acceleration of 6ms^-2 covers 45m in the 6th second of its motion.
Find its initial velocity.
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<pre>
Let u be the initial velocity.


Then the velocity at the end of the 5-th second is  u + 5a = u + 5*6 = u + 30 m/s;

     the velocity at the end of the 6-th second is  u + 6a = u + 6*6 = u + 36 m/s;

     the average velocity during the 6-th second is  {{{u[average]}}} = {{{((u+30) + (u+36))/2}}} = u + 33 m/s.


     an equation for the distance, covered during the 6-th second is

          {{{u[average]*1}}} = 45  meters,

     or

          (u+33) = 45,

     which gives

           u = 45 - 33 = 12 m/s.


<U>ANSWER</U>.  The initial velocity is 12 m/s.
</pre>

Solved.