Question

A particle is moving in a straight line and its distance s cm from a fixed point in the line after t seconds is given by s = 12t – 15t^2 + 4t^3. What will be the distance between the two positions of the particle at two times, when the velocity is instantaneously 0?

(a) 27/4 cm

(b) 29/4 cm

(c) 27/2 cm

(d) 29/2 cm

I have been asked this question at a job interview.

The above asked question is from Calculus Application topic in division Application of Calculus of Mathematics – Class 12

Answer 1

Correct answer is (a) 27/4 cm

The best I can explain: We have, s = 12t – 15t^2 + 4t^3 ……….(1)

Differentiating both side of (1) with respect to t we get,

(ds/dt) = 12 – 30t + 12t^2

Clearly, the velocity is instantaneously zero, when

(ds/dt) = 12 – 30t + 12t^2 = 0

Or 12 – 30t + 12t^2 = 0

Or (2t – 1)(t – 2) = 0

Thus, t = 2 or t = ½

Putting the value t = 2 and t = ½ in (1),

We get, when t = 2 then s = (s1) = 12(2) – 15(2)^2 + 4(2)^3 = -4.

When t = ½ we get, s = (s2) = 12(1/2) – 15(1/2)^2 + 4(1/2)^3 = 11/4.

Thus, the distance between the two positions of the particle at two times, when the velocity is instantaneously 0 = s2 – s1

= 11/4 – (-4)

= 27/4 cm.