Question

A machine moves an object by giving it a constant power P. If object being its motion from rest, then at any time t, its velocity will be proportional to:
1. √t
2. t^3/2
3. t
4. t²

Answer 1

Correct Answer - Option 1 : √t

CONCEPT:

Velocity: 

• It is defined as the rate of change of the object’s position with respect to a frame of reference and time.
  • It is a vector quantity.

​\(⇒ v = \frac{dx}{dt}\)

Power

• It is defined as the rate of doing work.

\(\Rightarrow P=\frac{W}{t}\)

CALCULATION:

Given P = constant and initial velocity u = 0 m/s

We know that,

\(\Rightarrow P=\frac{W}{t}\)     -----(1)

Where P = power, W = work done and t = time

• The work-done, when the force and the displacement are in the same direction is given as,

\(\Rightarrow W=Fs\)     -----(2)

Where F = force and s = displacement

• By the second equation of motion,

\(\Rightarrow s=ut+\frac{1}{2}at^{2}\)

\(\Rightarrow s=(0\times t)+\frac{1}{2}at^{2}\)

\(\Rightarrow s=\frac{1}{2}at^{2}\)     -----(3)

By Newton's second law of motion,

\(\Rightarrow F=ma\)     -----(4)

Where m = mass and a = acceleration

By equation 1, equation 2, equation 3, and equation 4,

\(\Rightarrow P=\frac{ma^{2}t}{2}\)     -----(5)

By the first equation of motion,

\(\Rightarrow v=u+at\)

\(\Rightarrow v=0+at\)

\(\Rightarrow v=at\)

\(\Rightarrow a=\frac{v}{t}\)     -----(6)

By equation 5 and equation 6,

\(\Rightarrow P=\frac{mv^{2}t}{2t^2}\)

\(\Rightarrow P=\frac{mv^{2}}{2t}\)

\(\Rightarrow v^2=\frac{2Pt}{m}\)

\(\Rightarrow v=\sqrt\frac{2Pt}{m}\)

Since P and m are constant,

\(\Rightarrow v\propto\sqrt t\)

• Hence, option 1 is correct.