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Chapter 2: Problem 61

A particle located at \(x=0\) at time \(t=0\), starts moving along the positive \(x\)-direction with a velocity \(v\) that varies as \(v=\alpha \sqrt{x}\). The displacement of the particle varies with time as (A) \(t^{3}\) (B) \(t^{2}\) (C) \(t\) (D) \(t^{1 / 2}\)

Short Answer

Expert verified
The displacement of the particle varies with time as \(x \propto t^2\), which corresponds to option (B). Therefore, the correct answer is: (B) \(t^2\).

Step by step solution

01

Write down the known variables

We are given that: 1. The particle is initially located at x = 0 and t = 0. 2. The velocity of the particle, v, varies as \(v = \alpha \sqrt{x}\). Our goal is to find the correct expression for x in terms of t.
02

Rewrite the velocity equation in terms of dx/dt

The velocity, v, is the first derivative of the displacement, x, with respect to time, t. Hence, recall that \(v = \frac{dx}{dt}\). We can rewrite the given equation as: \(\frac{dx}{dt} = \alpha \sqrt{x}\).
03

Separate the variables

We want to get x on one side of the equation and t on the other. To do this, we will multiply both sides of the equation by \(dt\) and divide both sides by \(\sqrt{x}\) to obtain: \(\frac{dx}{\sqrt{x}} = \alpha dt\).
04

Integrate both sides of the equation

Now, we will integrate both sides of the equation: \[\int \frac{dx}{\sqrt{x}} = \int \alpha dt\]. By evaluating the integrals, we get: \[\frac{dx}{\sqrt{x}} = \int \frac{dx}{x^{\frac{1}{2}}}\] \[2\sqrt{x} = \alpha t + C\], where C is the constant of integration.
05

Solve for the constant of integration using the initial conditions

Using the initial conditions that the particle is initially located at x = 0 and t = 0, we can find the constant of integration: \(2\sqrt{0} = \alpha(0) + C\) Thus, \(C = 0\).
06

Rewrite the equation to express x in terms of t

Now that we know C, we can write the equation expressing x in terms of t: \(2\sqrt{x} = \alpha t\) Square both sides of the equation: \(4x = \alpha^2 t^2\) Divide both sides by 4: \(x = \frac{\alpha^2}{4}t^2\) The displacement of the particle varies with time as \(x \propto t^2\), which corresponds to option (B). Therefore, the correct answer is: (B) \(t^2\).

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Most popular questions from this chapter

A particle is projected from ground with velocity \(40 \sqrt{2} \mathrm{~m} / \mathrm{s}\) at \(45^{\circ}\). At time \(t=2 \mathrm{~s}\) : (A) Displacement of particle is \(100 \mathrm{~m}\) (B) Vertical component of velocity is \(30 \mathrm{~m} / \mathrm{s}\) (2) with (C) Velocity makes an angle of \(\tan ^{-1}\) horizontal (D) Particle is at height of \(80 \mathrm{~m}\) from ground

A road is \(5 \mathrm{~m}\) wide. Its radius of curvature is \(20 \sqrt{6} \mathrm{~m}\). The outer edge is above the inner edge by a distance of \(1 \mathrm{~m}\). This road is most suited for a speed \(\left(g=10 \mathrm{~ms}^{-2}\right.\) ) (A) \(10 \mathrm{~ms}^{-1}\) (B) \(10 \sqrt{5} \mathrm{~ms}^{-1}\) (C) \(100 \mathrm{~ms}^{-1}\) (D) \(40 \sqrt{6} \mathrm{~ms}^{-1}\)

The position \((x)\) of a particle varies with time as \(x=\) \(t^{3}-3 t^{2}\), where \(t\) is time in second. Match Column-I with Column-II. $$ \begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \text { (A) Position of particle is } & \text { (1) } t=3 \mathrm{~s} \\\ \text { zero at } \\ \text { (B) Velocity of particle is } & \text { (2) } t=2 \mathrm{~s} \\ \text { zero at } \\ \text { (C) Magnitude of } & \text { (3) } t=5 \mathrm{~s} \\ \text { acceleration of particle } \\ \text { is zero at } \\ \begin{array}{ll} \text { (D) Magnitude of } \\ \text { instantaneous } \\ \text { acceleration is equal to } \\ 18 \mathrm{~m} / \mathrm{s}^{2} \end{array} & \text { (4) } t=1 \mathrm{~s} \\ \hline \end{array} $$

A particle is thrown horizontally from the top of a tower of height \(H .\) The angle made by velocity of particle before hitting the ground is \(45^{\circ}\) with the horizontal. What is the horizontal range of particle? (A) \(H\) (B) \(2 H\) (C) \(3 H\) (D) \(4 \mathrm{H}\)

A particle is projected with a velocity \(u\), at an angle \(\alpha\), with the horizontal. Time at which its vertical component of velocity becomes half of its net speed at the highest point will be (A) \(\frac{u}{2 g}\) (B) \(\frac{u}{2 g}(\sin \alpha-\cos \alpha)\) (C) \(\frac{u}{2 g}(2 \cos \alpha-\sin \alpha)\) (D) \(\frac{u}{2 g}(2 \sin \alpha-\cos \alpha)\)
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