A particle moves along the x-axis such that its time of travel t (in seconds) is given as a function of its position x (in meters) by the relation t = 4x^2 + 3x + 2 . The magnitude of the particle's acceleration when its velocity is 0.5 m/s is:

Step-by-step solution

Given the relation: t = 4x^2 + 3x + 2 Differentiating with respect to x : d t d x = 8x + 3 The velocity v is the reciprocal of d t d x : v = d x d t = 1 8x + 3 Acceleration a can be written using the chain rule as a = v d v d x . First, find d v d x : d v d x = d d x (8x + 3)⁻¹ = -1(8x + 3)⁻² 8 = -8v^2 Now, substitute this into the acceleration expression: a = v(-8v^2) = -8v^3 Given that v = 0.5 m/s (or 1 2 m/s), substitute this value: a = -8 ( 1 2 )^3 = -8 1 8 = -1 m/s ^2 The magnitude of the acceleration is 1 m/s