Question

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 distance from O after 4 seconds?

(a) 110 cm

(b) 120 cm

(c) 130 cm

(d) 140 cm

This question was addressed to me in an interview for job.

Origin of the question is Calculus Application topic in portion Application of Calculus of Mathematics – Class 12

Answer 1

Correct option is (b) 120 cm

Easy explanation: 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 and acceleration at time t seconds is, v = dx/dt and dv/dt respectively.

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)

Or ∫dx = ∫(5t + 3(t^2) + 4) dt

Or x = 5t^2/2 + t^3 + 4t + B   ……….(3)

By question x = 0, when t = 0;

Hence, from (3) we get, B = 0

Thus, x = 5t^2/2 + t^3 + 4t

Thus, distance of the particle after 4 seconds,

= [x]t = 4 = (5/2*4^2 + 4^3 + 4*4)  [putting t = 4 in (4)]

= 120 cm.