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?

Category: QuestionsA 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?
Editor">Editor Staff asked 11 months ago

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
 
I have been asked this question during an online interview.
 
This intriguing question comes from Calculus Application topic in portion Application of Calculus of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
Editor">Editor Staff answered 11 months ago

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.