A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds,respectively.What is the distance described by the particle in 3 seconds?

(a) 30 cm

(b) 31 cm

(c) 32 cm

(d) 33 cm

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This interesting question is from Calculus Application in chapter Application of Calculus of Mathematics – Class 12

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Right answer is (b) 31 cm

The best I can explain: We assume that the particle moves with uniform acceleration 2f m/sec.

Let, x m be the distance of the particle from a fixed point on the straight line at time t seconds.

Let, v be the velocity of the particle at time t seconds, then,

So, dv/dt = 2f

Or ∫dv = ∫2f dt

Or v = 2ft + b ……….(1)

Or dx/dt = 2ft + b

Or ∫dx = 2f∫tdt + ∫b dt

Or x = ft^2 + bt + a ……….(2)

Where, a and b are constants of integration.

Given, x = 21, when t = 2; x = 43, when t = 4 and x = 91, when t = 7.

Putting these values in (2) we get,

4f + 2b + a = 21 ……….(3)

16f + 4b + a = 43 ……….(4)

49f + 7b + a = 91 ……….(5)

Solving (3), (4) and (5) we get,

a = 7, b = 5 and f = 1

Therefore, from (2) we get,

x = t^2 + 5t + 7

Therefore, the distance described by the particle in 3 seconds,

= [x]t = 3 = (3^2 + 5*3 + 7)m = 31m