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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

Answer»

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)


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