Two particles start from rest from the same point and move along the same straight path; the first starts with a uniform velocity of 20 m/minute and the second with a uniform acceleration of 5 m/ min^2. Before they meet again, then what will be the maximum distance?

Category: QuestionsTwo particles start from rest from the same point and move along the same straight path; the first starts with a uniform velocity of 20 m/minute and the second with a uniform acceleration of 5 m/ min^2. Before they meet again, then what will be the maximum distance?
Editor">Editor Staff asked 11 months ago

Two particles start from rest from the same point and move along the same straight path; the first starts with a uniform velocity of 20 m/minute and the second with a uniform acceleration of 5 m/ min^2. Before they meet again, then what will be the maximum distance?
 
(a) 38m
 
(b) 36m
 
(c) 42m
 
(d) 40m
 
The question was asked during an internship interview.
 
I’m obligated to ask this question of Calculus Application in section 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 (d) 40m
 
To elaborate: Suppose, the two particle starts from rest at and move along the straight path OA.
 
Further assume that the distance between the particle is maximum after t minutes from start (before they meet again).
 
If we be the position of the particle after 30 minutes from the start which moves with uniform acceleration 5 m/min^2 and C that of the particle moving with uniform velocity 20 m/min, then we shall have,
 
OB = 1/2(5t^2) and OC = 20t
 
If the distance between the particle after t minutes from start be x m, then,
 
x = BC = OC – OB = 20t – (5/2)t^2 ……….(1)
 
Now, dx/dt = 20 – 5t and d^2x/dt^2 = -5
 
For maximum or minimum values of x, we have,
 
dx/dt = 0
 
Or 20 – 5t = 0
 
Or t = 4
 
And [d^2y/dx^2] = -5 < 0
 
Thus, x is maximum at t = 4.
 
Therefore, the maximum value is,
 
= 20*4 – (5/2)(4^2)  [putting t = 4 in (1)]
 
= 40 m