91Ó°ÊÓ

The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy: \[W = ΔK.E.\] Where \(W\) is the work done, \(ΔK.E.\) is the change in kinetic energy.

The kinetic energy of an object can be written as: \[K.E. = \frac{1}{2}mv^2\] where \(m\) is the mass of the object and \(v\) is its velocity.

Using the given information, we know that the distance moved is proportional to the power function of time, so let's write the change in kinetic energy as: \[ΔK.E. = \frac{1}{2}m(v_f^2 - v_i^2)\] Where \(v_f\) is the final velocity and \(v_i\) is the initial velocity of the body.

Power is defined as the rate at which work is done, which can be written as: \[P = \frac{W}{t}\] Where \(P\) is the power, \(W\) is the work done, and \(t\) is the time taken.

Since work done equals the change in kinetic energy, we can replace the work, \(W\), with the expression for the change in kinetic energy. We can now write the power equation as: \[P = \frac{ΔK.E.}{t} = \frac{\frac{1}{2}m(v_f^2 - v_i^2)}{t}\]

Since the power is constant, we can rewrite the equation with the distance variable, L, such that: \[\frac{dL}{dt} = kv^2\] Integrating both sides with respect to time: \(L(t) = k\int_{0}^{t}{v^2(t')dt'}\) We don't know the specific relationship between \(v(t)\) and \(t\), so we can't directly provide a single expression for how the distance relates to time. However, we can analyze the given options to find the correct one.

We are given the possible proportional relationships between distance and time: \[L(t) \propto t^{\frac{3}{4}}, t^{\frac{3}{2}}, t^{\frac{1}{4}}, t^{\frac{1}{2}}\] Since power and time relationship is given as a second derivative, we will differentiate each relationship twice and inspect how the exponent of time in each case relates to the power. 1. Differentiating \(L(t) \propto t^{\frac{3}{4}}\) two times gives \(\frac{3}{8}t^{\frac{-1}{4}}\) 2. Differentiating \(L(t) \propto t^{\frac{3}{2}}\) two times gives \(3t^{\frac{1}{2}}\) 3. Differentiating \(L(t) \propto t^{\frac{1}{4}}\) two times gives \(\frac{1}{8}t^{\frac{-3}{4}}\) 4. Differentiating \(L(t) \propto t^{\frac{1}{2}}\) two times gives \(\frac{1}{2}t^{\frac{-1}{2}}\) Now, we want a time-exponent increasing by 2 from the given options in the proportional distances since our relationship needs to relate to power (a second derivative). Only option 2 gives this: \[L(t) \propto t^{\frac{3}{2}}\] Therefore, the correct answer is: (B) \(t^{\frac{3}{2}}\)