Q126 / 180 · COMEDK 2021 · Physics
Which graph shows the correct $v-x$ graph of a freely falling body?
- Correct answerA
- B
- C
- D
Solution
For a body falling freely under gravity from rest, the initial velocity $u = 0$ and acceleration $a = g$.
The equation of motion is $v^2 - u^2 = 2as$.
Substituting $u = 0$, $a = g$, and displacement $s = x$, we get $v^2 = 2gx$.
This implies $v = \sqrt{2gx}$, which can be written as $v = \sqrt{2g} \cdot \sqrt{x}$.
This equation represents a parabolic relationship between $v$ and $x$, specifically of the form $v^2 \propto x$.
The graph of $v$ versus $x$ is a parabola opening along the positive $x$-axis, starting from the origin $(0,0)$.
Comparing this with the given options, the first graph shows this parabolic behavior starting from the origin.
Answer:
The equation of motion is $v^2 - u^2 = 2as$.
Substituting $u = 0$, $a = g$, and displacement $s = x$, we get $v^2 = 2gx$.
This implies $v = \sqrt{2gx}$, which can be written as $v = \sqrt{2g} \cdot \sqrt{x}$.
This equation represents a parabolic relationship between $v$ and $x$, specifically of the form $v^2 \propto x$.
The graph of $v$ versus $x$ is a parabola opening along the positive $x$-axis, starting from the origin $(0,0)$.
Comparing this with the given options, the first graph shows this parabolic behavior starting from the origin.
Answer:
Hence, the correct answer is option: 'A'.