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?

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

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
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience

1 Answers
Editor">Editor Staff answered 11 months ago

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.