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/sec2. What will be the velocity of the particle from O after 4 seconds?

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/sec2. What will be the velocity of the particle from O after 4 seconds? Correct Answer 72 cm/sec

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*(t2)/2 + A   ……….(1) By question v = 4, when t = 0; Hence, from (1) we get, A = 4. Thus, v = dx/dt = 5t + 3(t2) + 4  ……….(2) Thus, velocity of the particle after 4 seconds, = t = 4 = (5*4 + 3*42 + 4) = 72 cm/sec.