Question

A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time ‘t’ is proportional to
A. \[t\]
B. \[{t^{3/2}}\]
C. \[{t^{1/2}}\]
D. \[{t^{2/3}}\]

Answer 1

Hint: In this question, we need to find the proportionate factor for the distance traveled by box if a constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. For this, we need to use the following formulae.

Formula used:
The power of an object is the product of force and velocity.
Mathematically, it is denoted as \[P = F \times V\]
Also, the force of an object is the product of mass and acceleration.
Mathematically, it is denoted as \[F = m \times a\]

Complete step by step solution:
We know that there is constant power.
So, we get \[P = C\]
where, \[P\] is a power and \[C\] is constant.
Also, \[P = F \times V{\text{ }}......{\text{ }}\left( 1 \right)\]
But \[F = m \times a\] and \[a = \dfrac{{dv}}{{dt}}\] as acceleration is the rate of change of velocity.
Now, put these values in the equation (1)
Thus, we get
\[
  P = m \times a \times V \\
   \Rightarrow P = m \times \dfrac{{dv}}{{dt}} \times V \\
 \]
Let us simplify it further.
\[Pdt = mVdv\]
By taking integration on both sides, we get
\[
  \int\limits_0^t {Pdt} = \int\limits_0^v {mVdv} \\
   \Rightarrow \int\limits_0^t {Pdt} = m\int\limits_0^v {Vdv} \\
   \Rightarrow \left[ {Pt} \right]_0^t = m\left[ {\dfrac{{{V^2}}}{2}} \right]_0^v \\
   \Rightarrow \left[ {Pt - 0} \right]_0^t = m\left[ {\dfrac{{{V^2}}}{2} - 0} \right]_0^v \\
 \]
By simplifying further, we get
\[ \Rightarrow \left[ {Pt} \right] = \dfrac{{m{V^2}}}{2}\]
Now, find the value of \[V\]from the above equation.
\[ \Rightarrow {V^2} = \dfrac{{2Pt}}{m}\]
By taking square root on both sides, we get
\[ \Rightarrow V = \sqrt {\dfrac{{2Pt}}{m}} ......\left( 1 \right)\]
We know that the velocity is the rate of change of displacement.
So, \[V = \dfrac{{dx}}{{dt}}\]
Put this value of \[V\] in the equation (1)
Thus, we get
\[
  \dfrac{{dx}}{{dt}} = \sqrt {\dfrac{{2Pt}}{m}} \\
   \Rightarrow dx = \left( {\sqrt {\dfrac{{2Pt}}{m}} } \right)dt \\
 \]
By integrating on both sides, we get
\[
  \int {dx} = \int {\left( {\sqrt {\dfrac{{2Pt}}{m}} } \right)dt} \\
   \Rightarrow \int {dx} = \left( {\sqrt {\dfrac{{2P}}{m}} } \right)\dfrac{d}{{dt}}\left( {{{\sqrt t }^{}}} \right) \\
   \Rightarrow x = \left( {\sqrt {\dfrac{{2P}}{m}} } \right)\left( {\dfrac{{{t^{1 + \dfrac{1}{2}}}}}{{\dfrac{3}{2}}}} \right) + C \\
   \Rightarrow x = \dfrac{2}{3}\left( {\sqrt {\dfrac{{2P}}{m}} } \right)\left( {{t^{\dfrac{3}{2}}}} \right) + C \\
 \]
This indicates that \[x\] is directly proportional to the \[{t^{\dfrac{3}{2}}}\].

Therefore, the correct option is (B).

Note: Newton's second law says that the acceleration of an object is determined by two variables: the total force acting on the object and its mass. The acceleration of the body is related to the overall force applied to it and inversely proportional to its mass. Many students make mistakes in the calculation part of the integration. This is the only way through which we can solve this example in an easy manner.