The question describes a specific relationship between the distance covered by an object and the time elapsed. We need to determine the nature of the object's acceleration based on this relationship.

Motion Analysis: Distance Proportional to Time Squared

The problem states that the distance ($d$) covered by an object is in direct proportion to the square of the time elapsed ($t$). This relationship can be expressed mathematically as:

$d \propto t^2$

This proportionality means that the distance is equal to some constant value multiplied by the square of the time. We can write this as:

$d = k \cdot t^2$

Here, $k$ is the constant of proportionality.

Understanding Kinematic Equations for Motion

To understand the acceleration, we can relate this given condition to the standard kinematic equations used in physics to describe motion. One of the key equations that relates distance, initial velocity, time, and acceleration is:

$d = v_0 t + \frac{1}{2} a t^2$

In this equation:

• $d$ represents the distance traveled.
• $v_0$ represents the initial velocity of the object (velocity at time $t=0$).
• $t$ represents the time elapsed.
• $a$ represents the constant acceleration of the object.

Deriving the Conclusion on Acceleration

Let's consider the specific case often implied in such problems: the object starts from rest. If an object starts from rest, its initial velocity ($v_0$) is zero.

Substituting $v_0 = 0$ into the kinematic equation, we get:

$d = (0) t + \frac{1}{2} a t^2$

This simplifies to:

$d = \frac{1}{2} a t^2$

Now, let's compare this derived equation with the relationship given in the problem, $d = k \cdot t^2$.

By comparing the two equations:

$k \cdot t^2 = \frac{1}{2} a t^2$

We can see that the constant of proportionality $k$ is equal to $\frac{1}{2} a$.

Since the problem states $d$ is directly proportional to $t^2$, the value of $k$ is constant. If $k$ is constant, then $\frac{1}{2} a$ must also be constant.

Therefore, if $\frac{1}{2} a$ is constant, it means that the acceleration ($a$) of the object must be constant.

Evaluating the Options

Based on our analysis:

• Option 1 (Zero acceleration): If $a=0$, then $d = 0$ (assuming $v_0=0$), which does not fit the $d \propto t^2$ relationship unless $d$ is always zero.
• Option 2 (Constant acceleration): This matches our derivation ($a = \text{constant}$).
• Option 3 (Increasing acceleration): If acceleration increases with time, the distance covered would typically be proportional to a higher power of time (e.g., $t^3$), not $t^2$.
• Option 4 (Decreasing acceleration): If acceleration decreases with time, the distance covered would not follow the $d \propto t^2$ pattern.

Thus, the conclusion that can be drawn is that the object is moving with constant acceleration.