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?

Category: QuestionsA 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?
Editor">Editor Staff asked 11 months ago

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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1 Answers
Editor">Editor Staff answered 11 months ago

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