Question

If velocity of the particle is given by $v = \sqrt{x}$, where x denotes the position of the particle and initially particle was at x = 4m, then which of the following are correct.

• A. At t = 2 s, the position of the particle is at x = 9m
• B. Particle acceleration at t = 2 s is 1 m/s²
• C. Particle acceleration is 1/2 m/s² throughout the motion
• D. Particle will never go in negative direction from its starting position

Answer 1

Answer: (A), (C), (D)

A, C, D

Answer 2

The correct options are (A), (C), and (D).

1. Find position $x(t)$: We know $v = \frac{dx}{dt}$. So, $\frac{dx}{dt} = \sqrt{x}$ Separate variables: $\frac{dx}{\sqrt{x}} = dt$ Integrate from initial conditions $(x_0=4, t_0=0)$ to $(x, t)$: $\int_{4}^{x} x^{-1/2} dx = \int_{0}^{t} dt$ $[2\sqrt{x}]_{4}^{x} = [t]_{0}^{t}$ $2\sqrt{x} - 2\sqrt{4} = t$ $2\sqrt{x} - 4 = t$ $2\sqrt{x} = t + 4$ $\sqrt{x} = \frac{t+4}{2}$ $x(t) = \left(\frac{t+4}{2}\right)^2 = \frac{(t+4)^2}{4}$

2. Find acceleration $a(t)$: Acceleration $a = \frac{dv}{dt}$. Since $v = \sqrt{x}$, we can use the chain rule: $a = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$. Given $v = x^{1/2}$, then $\frac{dv}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$. So, $a = \sqrt{x} \left(\frac{1}{2\sqrt{x}}\right) = \frac{1}{2}$ m/s². The acceleration is constant throughout the motion.

3. Evaluate the options: (A) At $t = 2 s$, the position of the particle is at $x = 9m$. Substitute $t=2$ into $x(t)$: $x(2) = \frac{(2+4)^2}{4} = \frac{6^2}{4} = \frac{36}{4} = 9m$. This statement is correct.

   (B) Particle acceleration at $t = 2 s$ is $1 m/s^2$. The acceleration is constant $a = 1/2$ m/s². This statement is incorrect.

   (C) Particle acceleration is $1/2 m/s^2$ throughout the motion. As derived, $a = 1/2$ m/s². This statement is correct.

   (D) Particle will never go in negative direction from its starting position. The starting position is $x=4m$. The position function is $x(t) = \frac{(t+4)^2}{4}$. For any $t \ge 0$, $(t+4)^2$ is always positive, and thus $x(t)$ is always positive. The particle moves from $x=4m$ to larger positive values. This statement is correct.