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MHT CET202120 Sep 2021Morning ShiftMathematicsApplication of DerivativesActual

A particle is moving on a straight line. The distance S travelled in time t is given by S=a t^2+b t+6 . If the particle comes to rest after 4 seconds at a distance of 16 ~m . from the starting point, then the acceleration of the particle is.

Options

  1. A-3 4 ~m / sec ^2
  2. B-1 2 ~m / sec ^2
  3. C-1 ~m / sec ^2
  4. D-5 4 ~m / sec ^2

Correct answer

D. -5 4 ~m / sec ^2

Step-by-step solution

aligned & S = at t ^2+ bt +6 & V = dS dt =2 at + b and A = d ^2 ~S dt ^2 =2 a aligned When particle comes to rest, aligned & S=16, t =4, ~V =0 & 16= a (4)^2+ b (4)+6 & 16 a +4 ~b =10 (1) aligned Also 0=2 a (1)(2)+4 b =-8 a From (1) and (2), we get 16 a+4(8 a)=10 -16=10 a= -5 9 We have acceleration A =2 a =2 ( -5 8 )= -5 4 ~m / sec ^2

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