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Step-by-Step Solution
Step 1: Recognize the Physical Situation
A box is pulled by a machine that delivers constant power $P$. The box starts from rest and moves along a horizontal surface. We need to find how the distance covered $x$ depends on time $t$.
Step 2: Express Power in Terms of the Variables
Power ($P$) can be written in terms of force ($F$) and velocity ($v$):
$P = F \cdot v$.
Since $F = m a$ for a box of mass $m$ and acceleration $a$, we have:
$P = m a \, v$.
Step 3: Relate Power to Velocity
Acceleration $a$ is the time derivative of velocity, that is $a = \frac{dv}{dt}$. Therefore,
$P = m \, v \, \frac{dv}{dt}$.
This can be rearranged as:
$P = \frac{m \, v \, dv}{dt}$.
Step 4: Integrate to Find the Velocity as a Function of Time
Separate variables and integrate both sides over time from $0$ to $t$ and velocity from $0$ to $v$:
$\int_{0}^{t} P \, dt = m \int_{0}^{v} v \, dv$.
The left side gives $Pt$ (because $P$ is constant), and the right side is $\frac{1}{2} m v^{2}$. Hence:
$P t = \frac{1}{2} m v^{2}.$
Solving for $v$:
$v = \sqrt{\frac{2 P t}{m}}.$
Step 5: Express Distance Moved ($x$) in Terms of Time ($t$)
Velocity is the time derivative of position: $v = \frac{dx}{dt}$. Substituting our expression for $v$:
$\frac{dx}{dt} = \sqrt{\frac{2 P t}{m}}.$
Rearranging:
$dx = \sqrt{\frac{2 P t}{m}} \, dt.$
Integrate both sides with respect to $t$:
$\int dx = \int \sqrt{\frac{2 P t}{m}} \, dt.$
We find that $x$ is proportional to:
$x \propto t^{3/2}.$
Step 6: Conclude the Proportionality
Thus, the distance $x$ covered by the box in time $t$ is proportional to $t^{3/2}$.
Final Answer
The correct choice is $t^{3/2}$.
