A particle moves along the straight-line OX, starting from O with a velocity 4 cm/sec. At time t seconds its acceleration is (5 + 6t) cm/sec^2. What will be the velocity of the particle from O after 4 seconds?
(a) 70 cm/sec
(b) 71 cm/sec
(c) 72 cm/sec
(d) 73 cm/sec
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This intriguing question comes from Calculus Application topic in portion Application of Calculus of Mathematics – Class 12
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Correct answer is (c) 72 cm/sec
The best I can explain: Let, vcm/sec be the velocity and x cm be the distance of the particle from O and time t seconds.
Then the velocity of the particle at time t seconds is, v = dx/dt
By the question, dv/dt = 5 + 6t
Or dv = (5 + 6t) dt
Or ∫dv = ∫(5 + 6t) dt
Or v = 5t + 6*(t^2)/2 + A ……….(1)
By question v = 4, when t = 0;
Hence, from (1) we get, A = 4.
Thus, v = dx/dt = 5t + 3(t^2) + 4 ……….(2)
Thus, velocity of the particle after 4 seconds,
= [v]t = 4 = (5*4 + 3*4^2 + 4) [putting t = 4 in (2)]
= 72 cm/sec.