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Chapter 2: Problem 145

The position of a particle is given by $$ r=3.0 t \hat{i}-2.0 t^{2} \hat{j}+4.0 \hat{k} \mathrm{~m} $$ Where \(t\) is in seconds and the coefficients have the proper units for \(r\) to be in metres. What is the magnitude of velocity of the particle \(t=2.0 \mathrm{~s}\) ? (A) \(\sqrt{72} \mathrm{~m} / \mathrm{s}\) (B) \(\sqrt{41} \mathrm{~m} / \mathrm{s}\) (C) \(\sqrt{11} \mathrm{~m} / \mathrm{s}\) (D) None

Short Answer

Expert verified
The magnitude of the velocity of the particle at \(t=2.0 \, \mathrm{s}\) is \(\sqrt{73} \, \mathrm{m/s}\).

Step by step solution

01

Differentiate the position vector with respect to time

The given position vector is: \(r = 3.0t\hat{i} - 2.0t^2\hat{j} + 4.0\hat{k} \, \mathrm{m}\). To find the velocity vector, we need to differentiate the position vector with respect to time: \(v = \frac{dr}{dt}\). Differentiate each component of the position vector with respect to time: \(\frac{d}{dt}(3.0t\hat{i}) = 3.0\hat{i}\) \(\frac{d}{dt}(-2.0t^2\hat{j}) = -4.0t\hat{j}\) \(\frac{d}{dt}(4.0\hat{k}) = 0\) Now, combine these results to find the velocity vector: \(v = 3.0\hat{i} - 4.0t\hat{j}\).
02

Evaluate the velocity vector at t=2.0 s

Now, we need to find the velocity vector at t=2.0 s. Substitute the given value of t into the velocity vector: \(v_t = 3.0\hat{i} - 4.0(2.0)\hat{j} = 3.0\hat{i} - 8.0\hat{j}\).
03

Find the magnitude of the velocity vector at t=2.0 s

To compute the magnitude of the velocity vector at t=2.0 s, we can use the following formula: \(|v_t|= \sqrt{(3.0)^2 + (-8.0)^2}\). Now, compute the magnitude: \(|v_t|= \sqrt{9 + 64} = \sqrt{73}\). The magnitude of the velocity of the particle at t=2.0 s is \(\sqrt{73} \, \mathrm{m/s}\). Since the result is not among the given choices, the correct answer is: (D) None

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Position Vector
In kinematics, the position vector is a fundamental concept. It tells us the location of a particle in space as a function of time. Consider a three-dimensional space. Here, the position vector typically looks like this:
  • The vector usually has components along the x, y, and z axes.
  • Each component describes the particle's position along a particular axis.
The given position vector for our particle is:
\( r = 3.0t\hat{i} - 2.0t^2\hat{j} + 4.0\hat{k} \, \text{m} \).

Breaking it down:
  • The term \( 3.0t\hat{i} \) means the particle moves along the x-axis.
  • The term \( -2.0t^2\hat{j} \) implies the particle's movement in the negative y-direction.
  • The constant \( 4.0\hat{k} \) indicates a fixed position along the z-axis.
Understanding how these terms interact over time helps us visualize how the particle travels through space.
Velocity Vector
The velocity vector is crucial in understanding how fast and in which direction a particle is traveling at any given time. To find the velocity vector, we must differentiate the position vector with respect to time. This differentiation explains how the position changes as time progresses.

In our example, the position vector is expressed as \( r = 3.0t\hat{i} - 2.0t^2\hat{j} + 4.0\hat{k} \, \text{m} \).
  • Differentiating, we get \( v = \frac{dr}{dt} \).
  • For \( 3.0t\hat{i} \), the derivative is a constant velocity of \( 3.0\hat{i} \).
  • For \( -2.0t^2\hat{j} \), the derivative is \( -4.0t\hat{j} \), indicating acceleration in the y-direction because it still includes \( t \).
  • Since \( 4.0\hat{k} \) is a constant in the z-direction, its derivative is zero.
Thus, the velocity vector is \( v = 3.0\hat{i} - 4.0t\hat{j} \), showcasing how two components impact the particle's motion.
Differentiation
Differentiation is a mathematical process used to find the rate at which a quantity changes. In kinematics, it helps us transition from position to velocity, and further, to acceleration if required. It's essential for predicting how the motion evolves over time.
  • To differentiate \( 3.0t\hat{i} \), where t is the variable, you multiply the coefficient by the power of \( t \) and reduce the power by one. Thus, you get \( 3.0\hat{i} \) since the power of \( t \) is 1.
  • With \( -2.0t^2\hat{j} \), using the power rule, results in \( -4.0t\hat{j} \). The squared term becomes linear after differentiation, which signifies how quickly the position is accelerating away from its origin in the y-direction.
  • A constant, such as \( 4.0\hat{k} \), becomes zero, indicating no change, thus no impact on velocity or acceleration.
In our problem, applying differentiation allowed us to find the instantaneous velocity; a critical step in understanding the particle's motion at every point in time.

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Most popular questions from this chapter

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

From a canon mounted on a wagon at height \(H\) from ground, a shell is fired horizontally with a velocity \(v_{0}\) with respect to canon. The canon and wagon has combined mass \(M\) and can move freely on the horizontal surface. The horizontal distance between shell and canon when the shell touches the ground is (A) \(v_{0} \sqrt{\frac{2 H}{g}}\) (B) \(\frac{v_{0} m}{M+m} \sqrt{\frac{2 H}{g}}\) (C) \(\frac{v_{0} M}{M+m} \sqrt{\frac{2 H}{g}}\) (D) \(\frac{v_{0} m}{M} \sqrt{\frac{2 H}{g}}\)

A jeep runs around a curve of radius \(0.3 \mathrm{~km}\) at a constant speed of \(60 \mathrm{~ms}^{-1}\). The jeep covers a curve of \(60^{\circ}\) arc. (A) Resultant change in velocity of jeep is \(60 \mathrm{~ms}^{-1}\). (B) Instantaneous acceleration of jeep is \(12 \mathrm{~ms}^{-1}\). (C) Average acceleration of jeep is \(11.5 \mathrm{~ms}^{-1}\). (D) Instantaneous and average acceleration are same in this case.

A particle is projected with a velocity \(v\) such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where \(g\) is acceleration due to gravity) (A) \(\frac{4 v^{2}}{5 g}\) (B) \(\frac{4 g}{5 v^{2}}\) (C) \(\frac{v^{2}}{g}\) (D) \(\frac{4 v^{2}}{\sqrt{5} g}\)

A particle is moving eastwards at a velocity of \(5 \mathrm{~ms}^{-1}\). In \(10 \mathrm{~s}\) the velocity changes to \(5 \mathrm{~ms}^{-1}\) northwards. The average acceleration in this time is (A) \(\frac{1}{2} \mathrm{~ms}^{-2}\) towards north \([2005]\) (B) \(\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2}\) towards north-east (C) \(\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2}\) towards north-west (D) Zero
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