Question

A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to:

• A. $ {{t}^{1/2}} $
• B. $ {{t}^{3/2}} $
• C. $ {{t}^{2}} $
• D. $ {{t}^{3/4}} $

Answer 1

The Correct Option is B

To solve this problem, we need to understand how constant power delivered by a machine affects the distance moved by a body over time.

When a machine delivers constant power \( P \) to a moving body, power is defined as:

P = \frac{dW}{dt}

where \( W \) is the work done. Work done \( W \) is also given by the product of force \( F \) and distance \( s \), so:

W = F \cdot s

Using the definition of power:

P = F \cdot v

where \( v \) is the velocity. Rearranging gives us:

v = \frac{P}{F}

Newton’s second law relates force to acceleration as:

F = m \cdot a

Thus, we can express velocity \( v \) as:

v = \frac{P}{m \cdot a}

At constant power, the kinetic energy \( KE \) is proportional to work done:

\frac{1}{2}mv^2 = P \cdot t

Rearranging gives:

v^2 = \frac{2P \cdot t}{m}

This means:

v \propto \sqrt{t}

Velocity is also defined as the rate of change of distance over time:

v = \frac{ds}{dt}

Substituting \( v \propto \sqrt{t} \) into the equation gives:

\frac{ds}{dt} \propto \sqrt{t}

Integrating both sides with respect to time \( t \), we get:

s \propto \int \sqrt{t} \, dt \propto t^{3/2}

Therefore, the distance moved by the body in time \( t \) is proportional to t^{3/2}.

The correct answer is t^{3/2}.