If time be a quadratic function of the length of the path of a moving point, Prove that the harmonic mean of the initial and final velocities is equal to the velocity at the middle point of the path, and that the tangential retardation is proportional to the cube of the velocity?

Question

Answer

Proof:

Harmonic Mean of Initial and Final Velocities:
- Let the time taken for the point to move from the initial velocity (v1) to the final velocity (v2) be T.
- Since time is a quadratic function of the length of the path, we know that time, T = k√s, where k is a constant.
- The velocity at the midpoint of the path is given by the formula: v = (2s/T) = (2s/(k√s)) = 2/k.
- The harmonic mean of v1 and v2 is given by: 2/((1/v1 + 1/v2)/2) = 2/(2/(v1 + v2)) = (v1 + v2)/2.
- Therefore, the harmonic mean of the initial and final velocities is equal to the velocity at the middle point of the path.

Tangential Retardation Proportional to Cube of Velocity:
- The tangential retardation (a) is given by the formula: a = dv/dt.
- From the given time function, we have T = k√s, which can be rearranged to s = (T^2)/(4k^2).
- Differentiating s with respect to time, we get v = ds/dt = (T/2k^2) ds/dT.
- Substituting for ds/dT from the time function, we get v = (T/2k^2) * (1/(2√s)) = T/(4k^2√s).
- Since a = dv/dt, the retardation a is proportional to the cube of velocity v^3, i.e., a ∝ v^3.

Therefore, we have proved that the harmonic mean of initial and final velocities is equal to the velocity at the middle point of the path, and the tangential retardation is proportional to the cube of the velocity.

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