A particle moves along a straight line such that the time t and position x are related by the equation t = 3x^2 - 4x . If v represents the instantaneous velocity of the particle, which of the following expressions gives its acceleration?

Step-by-step solution

Given t = 3x^2 - 4x Differentiating with respect to x : dt dx = 6x - 4 The instantaneous velocity v is given by: v = dx dt = 1 6x - 4 The acceleration a is given by: a = dv dt = v dv dx Differentiating v with respect to x : dv dx = d dx (6x - 4)⁻¹ = -1(6x - 4)⁻² 6 = -6(6x - 4)⁻² Since v = (6x - 4)⁻¹ , we have (6x - 4)⁻² = v^2 dv dx = -6v^2 Substituting this into the expression for acceleration: a = v(-6v^2) = -6v^3 Answer: -6v^3