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Chapter 2: Problem 53

A particle moves along \(x\)-axis as \(x=4(t-2)+a(t-2)^{2}\). Which of the following is true? (A) The initial velocity of particle is 4 (B) The acceleration of particle is \(2 a\) (C) The particle is at origin at \(t=0\) (D) None of these

Short Answer

Expert verified
The correct answer is choice (B): "The acceleration of the particle is \(2 a\)."

Step by step solution

01

Find the first derivative of the position function (Velocity function)

First, let's find the first derivative of the position function, which corresponds to the velocity function \(v(t)\). We will differentiate \(x(t)=4(t-2)+a(t-2)^{2}\) with respect to \(t\). Using the power rule of differentiation, we have: \(v(t) = \frac{dx}{dt} = 4 + 2a(t-2)\).
02

Evaluate the initial velocity of the particle

Now, let's find the initial velocity of the particle, which is the value of the velocity function when \(t=0\): \(v(0) = 4 + 2a(0-2) = 4 - 4a\). Now, let's check if this is equal to the value given in option A: \(v(0) = 4 - 4a \overset{?}{=} 4\). From this expression, we can see that the initial velocity of the particle is 4 if and only if \(a=0\). However, we cannot assume that \(a=0\), so the statement in option (A) is not true.
03

Find the second derivative of the position function (Acceleration function)

Now, let's find the second derivative of the position function, which corresponds to the acceleration function \(a(t)\). We will differentiate the velocity function \(v(t) = 4 + 2a(t-2)\) with respect to \(t\). Using the power rule of differentiation, we have: \(a(t) = \frac{d^2x}{dt^2} = 2a\). Since the second derivative is a constant value (doesn't depend on time), it means that the acceleration of the particle is constant and equal to \(2a\). Therefore, option (B) is true.
04

Evaluate the position of the particle at \(t=0\)

Finally, let's find the value of the position function at \(t=0\). We plug \(t=0\) into the given position function: \(x(0)=4(0-2)+a(0-2)^{2}= -8 + 4a\). Now, let's check if this is equal to the origin: \(x(0) = -8 + 4a \overset{?}{=} 0\). From this expression, we can see that the position of the particle at \(t=0\) is at the origin if and only if \(a=2\). However, we cannot assume that \(a=2\), so the statement in option (C) is not true. Since we've proven that option (B) is true and options (A) and (C) are not true, the correct answer is choice (B): "The acceleration of the particle is \(2 a\)."

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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

For a particle in uniform circular motion, the acceleration \(\vec{a}\) at a point \(\mathrm{P}(R, \theta)\) on the circle of radius \(R\) is (Here \(\theta\) is measured from the \(x\)-axis) (A) \(-\frac{v^{2}}{R} \cos \theta \hat{i}+\frac{v^{2}}{R} \sin \theta \hat{j}\) (B) \(-\frac{v^{2}}{R} \sin \theta \hat{i}+\frac{v^{2}}{R} \cos \theta \hat{j}\) (C) \(-\frac{v^{2}}{R} \cos \theta \hat{i}-\frac{v^{2}}{R} \sin \theta \hat{j}\) (D) \(\frac{v^{2}}{R} \hat{i}+\frac{v^{2}}{R} \hat{j}\)

A particle is projected upwards with a velocity of \(100 \mathrm{~m} / \mathrm{s}\) at an angle of \(37^{\circ}\) with the vertical. The time when the particle will move perpendicular to its initial direction is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}, \tan 53^{\circ}=4 / 3\right)\) (A) \(10 \mathrm{~s}\) (B) \(12.5 \mathrm{~s}\) (C) \(15 \mathrm{~s}\) (D) \(16 \mathrm{~s}\)

A particle can be projected with a given speed in two possible ways so as to make it pass through a point at a distance \(r\) from the point of projection. The product of the times taken to reach this point in the two possible ways is then proportional to (A) \(r\) (B) \(\frac{1}{r}\) (C) \(\frac{1}{r^{2}}\) (D) \(\frac{1}{r^{3}}\)

A particle has an initial velocity of \(3 \hat{i}+4 \hat{j}\) and an acceleration of \(0.4 \hat{i}+0.3 \hat{j}\). Its speed after \(10 \mathrm{~s}\) is [2009] (A) \(7 \sqrt{2}\) units (B) 7 units (C) \(8.5\) units (D) 10 units
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