A particle moving along a straight line with uniform acceleration, passes successively through three points A, B, C. If t1 = time taken to go from A to B; t2 = time taken to go from B to C and AB = a, BC = b, then what is the value of acceleration?

Category: QuestionsA particle moving along a straight line with uniform acceleration, passes successively through three points A, B, C. If t1 = time taken to go from A to B; t2 = time taken to go from B to C and AB = a, BC = b, then what is the value of acceleration?
Editor">Editor Staff asked 11 months ago

A particle moving along a straight line with uniform acceleration, passes successively through three points A, B, C. If t1 = time taken to go from A to B; t2 = time taken to go from B to C and AB = a, BC = b, then what is the value of acceleration?
 
(a) 2(bt1 – at2)/t1t2(t1 + t2)
 
(b) -2(bt1 – at2)/t1t2(t1 + t2)
 
(c) 2(bt1 + at2)/t1t2(t1 + t2)
 
(d) 2(bt1 – at2)/t1t2(t1 – t2)
 
I have been asked this question in an interview for internship.
 
Enquiry is from Calculus Application topic in chapter 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) 2(bt1 – at2)/t1t2(t1 + t2)
 
Best explanation: Let the particle beam moving with uniform acceleration f and its velocity at A be u.
 
Then, the equation of the motion of the particle from A to B is,
 
ut1 + ft1^2/2 = a   [as, AB = a]  ……….(1)
 
Again, the equation of motion of the particle from A to C is,
 
u(t1 + t2) + f(t1 + t2)^2/2 = a + b  [as, AC = AB + BC = a + b]  ……….(2)
 
Multiplying (1) by (t1 + t2) and (2) by t1 we get,
 
ut1 (t1 + t2) + ft1^2(t1 + t2)/2 = a(t1 + t2)  ……….(3)
 
And ut1(t1 + t2) + f(t1 + t2)^2/2 = (a + b)t1 ……….(4)
 
Subtracting (3) and (4) we get,
 
1/2(ft1)(t1 + t2)(t1 – t1 – t2) = at2 – bt1
 
Solving the above equation, we get,
 
f = 2(bt1 –  at2)/t1t2(t1 + t2)