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Q5.9:
A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time (t) is proportional to

Solution :

Step 1: Identification of Variables and Kinematic Relations

Let the body of mass $m$ be initially at rest at $t = 0$. The body undergoes one-dimensional motion under a constant acceleration $a$. Based on the first equation of motion [Per Kinematic laws for constant acceleration]:

$v(t) = u + at$

Since the body is initially at rest, $u = 0$. Thus, the velocity at any time $t$ is:

$v(t) = at$

Step 2: Force Formulation

According to Newton's Second Law of Motion, the net force $F$ acting on the body is the product of its mass $m$ and its acceleration $a$ [Per Newton's Second Law, $F = ma$]:

$F = ma$

Since $a$ is constant, the force $F$ is also constant throughout the duration of the motion.

Step 3: Deriving Power as a Function of Time

Power $P$ is defined as the dot product of force and velocity, or for one-dimensional motion, the product of the magnitudes [Per the definition of instantaneous power $P = \vec{F} \cdot \vec{v}$]:

$P(t) = F \cdot v(t)$

Substituting the expressions derived in Step 1 and Step 2:

$P(t) = (ma) \cdot (at)$

$P(t) = ma^2t$

Step 4: Analyzing Proportionality

From the expression $P(t) = (ma^2)t$, we observe the relationship between Power ($P$) and time ($t$). Since $m$ (mass) and $a$ (acceleration) are constants, the term $(ma^2)$ is a constant factor.

$P \propto t$

Visual Representation of Power Development

t P P ∝ t

The linear relationship indicates that for a body starting from rest under constant acceleration, the instantaneous power delivered increases linearly with time.

Final Solution: The power delivered to the body at time (t) is proportional to $t^1$ (i.e., directly proportional to time $t$).


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