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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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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 + 33 m/s.
an equation for the distance, covered during the 6-th second is
= 45 meters,
or
(u+33) = 45,
which gives
u = 45 - 33 = 12 m/s.
ANSWER. The initial velocity is 12 m/s.
Solved.