The position x of a particle moving along a straight line depends on time t according to the relation t = 3x^2 + 4x , where x is in meters and t is in seconds. If the magnitude of the acceleration of the particle when its velocity is 0.5 m s ⁻¹ is N 10⁻² m s ⁻² , the value of N is _____.

Step-by-step solution

Given the relation between time and position: t = 3x^2 + 4x Differentiating with respect to x , we get: dt dx = 6x + 4 The velocity v is the reciprocal of dt dx : v = dx dt = 1 6x + 4 The acceleration a is given by a = v dv dx . First, find dv dx : dv dx = d dx (6x + 4)⁻¹ = -1(6x + 4)⁻² 6 = -6v^2 Now, substitute this into the acceleration formula: a = v(-6v^2) = -6v^3 Given that the velocity v = 0.5 m s ⁻¹ = 1 2 m s ⁻¹ , the acceleration is: a = -6 ( 1 2 )^3 = - 6 8 = -0.75 m s ⁻² The magnitude of acceleration is 0