A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t^4/12 – 2t^3/3 + 3t^2/2 + t + 15. What is the minimum velocity?

Category: QuestionsA particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t^4/12 – 2t^3/3 + 3t^2/2 + t + 15. What is the minimum velocity?
Editor">Editor Staff asked 11 months ago

A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t^4/12 – 2t^3/3 + 3t^2/2 + t + 15. What is the minimum velocity?
 
(a) 1 cm/sec
 
(b) 2 cm/sec
 
(c) 3 cm/sec
 
(d) 4 cm/sec
 
I had been asked this question in my homework.
 
I need to ask this question from Calculus Application 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

The correct choice is (a) 1 cm/sec
 
The best explanation: Assume that the velocity of the particle at time t second is vcm/sec.
 
Then, v = dx/dt = 4t^3/12 – 6t^2/3 + 6t/2 + 1
 
So, v = dx/dt = t^3/3 – 2t^2/ + 3t + 1
 
Thus, dv/dt = t^2 – 4t + 3
 
And d^2v/dt^2 = 2t – 4
 
For maximum and minimum value of v we have,
 
dv/dt = 0
 
Or t^2 – 4t + 3 = 0
 
Or (t – 1)(t – 3) = 0
 
Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3
 
Now, [d^2v/dt^2]t = 3 = 2*3 – 4 = 2 > 0
 
Thus, v is minimum at t = 3.
 
Putting t = 3 in (1) we get,
 
3^3/3 – 2(3)^2/ + 3(3) + 1
 
= 1 cm/sec.