Question

Displacement (s) of a particle is related to time (t)

as:

2 3 s at bt ct

, where a, b and c are

constants of the motion. The velocity of the

particle when its acceleration is zero is given by​

Answer 1

Given :-

s = at + bt^2 - ct^3

accleration = 0

Solution:-

∴ v = dx/dt

 v = d(at + bt^2 -  ct^3)/dt

  v = a + 2bt - 3ct ------------(1)

∴ accleration = d^2x/dt^2

       a = 0 + 2b _- bct

as , accl^n = 0

so ,

 2b - bct = 0

t = d/3c

putting value of t in equation (1)

we get ,

V = a + 2b * b/3c - 36 * b^2/ 9c^2

V = a + 2b^2/3c - b^2/3c

therfore :-

V = a + b^2/3c

thank you

Answer 2

Answer:

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