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 velocity of the particle in 3 seconds?

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 velocity of the particle in 3 seconds? Correct Answer 11m/sec

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 = ft2 + 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 = t2 + 5t + 7 Putting t = 3, f = 1 and b = 5 in (1), We get, the velocity of the particle in 3 seconds, = t = 3 = (2*1*3 + 5)m/sec = 11m/sec.