NEET Physics Distance & Displacement MCQ Question

To analyze the relationship between initial velocity $v_0$, final velocity $v$, and displacement $x$ in the context of uniform acceleration, we can start by understanding the velocity-time graph and the area under the curve.

Explanation of the Correct Answer (C)

In the case of uniform acceleration, the velocity of the object changes linearly over time. The velocity-time graph will therefore be a straight line. The initial velocity is represented by $v_0$ at $t=0$ and the final velocity $v$ at time $t=T$.

The area under the velocity-time graph from $t=0$ to $t=T$ gives us the total displacement $x$ of the object:

$$x = \text{Area under the curve from } t=0 \text{ to } t=T$$

Since the graph is a trapezium (or a triangle if initial velocity is zero), we can find the area using the formula for the area of a trapezoid:

$$\text{Area} = \frac{1}{2} \times (b_1 + b_2) \times h$$

Where:

• $b_1 = v_0$ (initial velocity)
• $b_2 = v$ (final velocity)
• $h = T$ (time duration)

Thus, the displacement $x$ can be expressed as:

$$x = \frac{1}{2} \times (v_0 + v) \times T$$

This formula shows that the displacement $x$ is equal to the arithmetic average of the initial and final velocities multiplied by the time $T$. This confirms that option C is correct.

Clarification of Other Options

Option A: "The displacement $x$ is equal to the area of a triangle formed by $v_0$, $v$, and $T$."

This option is incorrect because the area under the curve is not just a triangle; it is a trapezoid (or a combination of a rectangle and triangle if $v_0$ is not zero).

Option B: "The displacement $x$ is equal to the area of a rectangle formed by $v_0$ and $T$."

This is incorrect because the rectangle would only represent the displacement if the velocity were constant at $v_0$ for the entire time $T$. However, in the case of uniform acceleration, the velocity changes from $v_0$ to $v$.

Option D: "The displacement $x$ is solely determined by the final velocity $v$ and time $T$."

This is incorrect because displacement also depends on the initial velocity $v_0$. The effect of both initial and final velocities is necessary to calculate accurate displacement under uniform acceleration.

Summary of Key Formulas

1. Displacement under uniform acceleration:

$$x = \frac{1}{2} (v_0 + v) T$$

1. Final velocity in terms of initial velocity, acceleration, and time:

$$v = v_0 + aT$$

1. Average velocity under uniform acceleration:

\text{Average velocity} = \frac{v_0 + v}{2}$$ By understanding these relationships and the geometric interpretation of the velocity-time graph, we can confidently conclude that the correct answer is **C**.