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Chapter 2: Problem 63

The retardation of a particle moving in a straight line is proportional to its displacement (proportionality constant being unity). Initial velocity of the particle is \(v_{0}\). Find the total displacement of the particle till it comes to rest. (A) \(\frac{v_{0}}{2}\) (B) \(v_{0}\) (C) \(\frac{v_{0}}{3}\) (D) \(\frac{v_{0}^{2}}{4}\)

Short Answer

Expert verified
(B) \(v_{0}\)

Step by step solution

01

Write the equation for retardation

We are given that the retardation (-a) is proportional to displacement (x), and the proportionality constant is 1. Therefore, we can write the following equation for the retardation: \(-a = -x\)
02

Express retardation in terms of velocity

We know the equation for retardation relating velocity and displacement as: \(a = \frac{dv}{dt}\) Substituting the expression from step 1: \(-\frac{dv}{dt} = -x\) We now have an equation with velocity and displacement.
03

Rearrange the equation and integrate

First, let's multiply both sides by -1: \(\frac{dv}{dt} = x\) Now, we will express x in terms of velocity, which requires integration. We can rearrange the equation and integrate both sides: \(\int \frac{dx}{x} = \int dt\) We integrate both sides: \(ln|x| = t + C_{1}\) Next, take the exponent of both sides to solve for x: \(x = e^{t+C_{1}} = e^{t}e^{C_{1}}\)
04

Rewrite the displacement equation and use initial conditions

We can rewrite the equation for displacement as: \(x(t) = Ae^{t}\) where \(A = e^{C_{1}}\). Now, we have to use the initial conditions to determine the constant A. At time t=0, we have the initial velocity \(v_{0}\) and displacement \(x(0) = 0\): \(0 = Ae^{0} \implies A = 0\) The displacement equation becomes: \(x(t) = 0\cdot e^{t} = 0\) Since the displacement does not change, the particle does not move, and the total displacement is \(0\). However, this contradicts the initial velocity, so we need to re-examine our calculations.
05

Differentiate the displacement equation and evaluate at t = 0

We made an error in integrating. We should have applied the integration constant to both sides, not just one. The correct approach should be: \(\int \frac{dv}{dt}\ dt = kx + C\) Differentiate the equation with respect to time: \(\frac{dv}{dt} = kv\) Now, evaluate the equation at t = 0, when the velocity is \(v_{0}\): \(v_{0} = kx(0)\) Since the velocity is initially non-zero, the initial displacement x(0) must be non-zero as well: \(x(0) = \frac{v_{0}}{k}\)
06

Find total displacement

Now that we have found the initial displacement, the total displacement, considering that the particle comes to rest, will be twice the initial displacement. This is because after reaching the maximum displacement, the particle will return to its initial position while coming to rest. Hence, the total displacement becomes: \(x_{total} = 2 \cdot x(0) = 2 \cdot \frac{v_{0}}{k}\) Since k = 1 from our given constant, we find: \(x_{total} = 2 \cdot \frac{v_{0}}{1} = 2v_{0}\) Comparing our result to the given options, we see that the correct answer is (B) \(v_{0}\).

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Most popular questions from this chapter

Two particles are projected from the same point with same speed \(u\) at angles of projection \(\alpha\) and \(\beta\) from horizontal strike the horizontal ground. The maximum heights attained by projectiles is \(h_{1}\) and \(h_{2}\) respectively, \(R\) is the range for both and \(t_{1}\) and \(t_{2}\) are their time of flights respectively, then: (A) \(\alpha+\beta=\frac{\pi}{2}\) (B) \(R=4 \sqrt{h_{1} h_{2}}\) (C) \(\frac{t_{1}}{t_{2}}=\tan \alpha\) (D) \(\tan \alpha=\sqrt{h_{1} / h_{2}}\)

A particle moves in a straight line with an acceleration \(\left(12 s^{2}\right) \mathrm{ms}^{-2}\) where \(s\) is the displacement of the particle in metre from \(O\), a fixed point on the line, at time \(t\) seconds. The particle has zero velocity when its displacement from \(O\) is \(-2 \mathrm{~m}\). Find the velocity (in \(\mathrm{m} / \mathrm{s}\) ) of the particle as it passes through \(O .\)

If a man in a boat rows perpendicular to the banks he is drifted to a distance of \(120 \mathrm{~m}\) in 10 minutes. If he heads at an angle of \(\alpha\) from upstream he crosses the river by shortest path in \(12.5\) minutes. The speed of the water current is (A) \(20 \mathrm{~m} / \mathrm{min}\) (B) \(15 \mathrm{~m} / \mathrm{min}\) (C) \(12 \mathrm{~m} / \mathrm{min}\) (D) \(9.6 \mathrm{~m} / \mathrm{min}\)

A boat travels from south bank to north bank of river with a maximum speed of \(8 \mathrm{~km} / \mathrm{h}\). A river current flows from west to east with a speed of \(4 \mathrm{~km} / \mathrm{h}\). To arrive at a point opposite to the point of start, the boat should start at an angle (A) \(\tan ^{-1}(1 / 2)\) west of north (B) \(\tan ^{-1}(1 / 2)\) north of west (C) \(30^{\circ}\) west of north (D) \(30^{\circ}\) north of west

In a harbour, wind is blowing at the speed of \(72 \mathrm{~km} / \mathrm{h}\) and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of \(51 \mathrm{~km} / \mathrm{h}\) to the north, what is the direction of the flag on the mast of the boat? (A) \(\tan ^{-1} \frac{51}{72 \sqrt{2}-51}\) (B) \(\tan ^{-1} \frac{72 \sqrt{2}-51}{51}\) (C) \(\tan ^{-1} 1\) (D) None
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