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Chapter 3: Problem 45

A particle's position is \(\vec{r}=\left(c t^{2}-2 d t^{3}\right) \hat{\imath}+\left(2 c t^{2}-d t^{3}\right) \hat{\jmath}\) where \(c\) and \(d\) are positive constants. Find expressions for times \(t>0\) when the particle is moving in (a) the \(x\) -direction and (b) the \(y\) -direction.

Short Answer

Expert verified
The time at which the particle is moving purely in the x-direction is \(t=\frac{4c}{3d}\) and the time at which the particle is moving purely in the y-direction is \(t=\frac{c}{3d}\).

Step by step solution

01

Calculating Velocity

The velocity vector of the particle is the derivative of the position vector, which can be found as \( \vec{v} = \frac{d\vec{r}}{dt} = (2ct - 6dt^2)\hat{\imath} + (4ct-3dt^2)\hat{\jmath}\)
02

Find when velocity along y is 0

In order to find out when the particle is only moving along the x direction, the y-component of the velocity vector must be zero. This will imply the motion is sideways only. Therefore, solve the equation \(4ct-3dt^2 = 0\) for \(t\). From this equation, we get two possible values: \(t=0\) or \(t= \frac{4c}{3d}\). However, we are asked to find the time \(t > 0\), so the answer to this part is \(t= \frac{4c}{3d}\).
03

Find when velocity along x is 0

For the particle to be moving in the y direction, the x-component of the velocity vector must be zero. Thus, solve the equation \(2ct - 6dt^2 = 0\) for \(t\). From this equation, we get two possible answers: \(t=0\) or \(t=\frac{c}{3d}\). Since we are interested in the time \(t > 0\), the answer to this part is \(t=\frac{c}{3d}\).

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Key Concepts

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

Velocity Vector
The velocity vector represents the rate of change of the particle's position with respect to time. In mathematics, it is the derivative of the position vector with respect to time.
  • For a given position vector \( \vec{r} \), the velocity vector \( \vec{v} \) is calculated by taking the derivative \( \frac{d\vec{r}}{dt} \).
  • This vector tells us how fast and in which direction the particle is moving.
  • In this context, the velocity vector is split into its x and y components, represented by \( (2ct - 6dt^2)\hat{\imath} + (4ct-3dt^2)\hat{\jmath} \).
Since velocity is a vector, any change in direction or speed introduces the need to analyze these components separately, as highlighted in the steps of the original solution.
Position Vector
The position vector defines the exact location of the particle in a coordinate system at any given point in time.
  • For the particle in question, the position vector is given by \( \vec{r} = (c t^{2}-2 d t^{3}) \hat{\imath} + (2 c t^{2} - d t^{3}) \hat{\jmath} \).
  • The vector is expressed in terms of unit vectors \( \hat{\imath} \) and \( \hat{\jmath} \), which define the direction in the coordinate plane.
  • The terms \( ct^2 \) and \( dt^3 \) in each component describe how the position changes over time based on constants \(c\) and \(d\).
Understanding the position vector is crucial as it lays the groundwork for calculating the velocity vector and determining the particle's motion.
Component Analysis
Component analysis breaks down vectors into their individual directional parts. For the particle's motion, it's essential to consider the horizontal and vertical components separately.
  • Each vector, like the velocity or position vector, is typically composed of an x-component and a y-component.
  • In the velocity vector \( (2ct - 6dt^2)\hat{\imath} + (4ct-3dt^2)\hat{\jmath} \), \( (2ct - 6dt^2) \) represents the x-component and \( (4ct-3dt^2) \) represents the y-component.
  • By setting either component to zero, we can find when the motion is restricted to one direction: purely along the x-axis or the y-axis.
This method simplifies complex problems and is especially useful in physics where understanding the direction of motion is as important as the motion itself.
Derivatives in Physics
Derivatives are a foundational element in physics, allowing for the analysis of changing quantities.
  • The derivative signifies how a quantity changes over time. In mechanics, it's often used to compute velocity or acceleration from position.
  • By deriving the position vector, we obtain the velocity vector, and further differentiating the velocity would give us acceleration.
  • In this exercise, the derivative of \( \vec{r} \) with respect to \( t \) provides essential information on how the particle's velocity evolves.
Thus, mastering derivatives offers a deeper understanding of motion and dynamics, making it a critical skill in both mathematics and physics.

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Most popular questions from this chapter

An ion in a mass spectrometer follows a semicircular path of radius \(15.2 \mathrm{cm} .\) What are (a) the distance it travels and (b) the magnitude of its displacement?

An object undergoes acceleration \(2.3 \hat{\imath}+3.6 \hat{\jmath} \mathrm{m} / \mathrm{s}^{2}\) for \(10 \mathrm{s}\). At the end of this time, its velocity is \(33 \hat{\imath}+15 \hat{\jmath} \mathrm{m} / \mathrm{s} .\) (a) What was its velocity at the beginning of the 10 -s interval? (b) By how much did its speed change? (c) By how much did its direction change? (d) Show that the speed change is not given by the magnitude of the acceleration multiplied by the time. Why not?

A migrating whale follows the west coast of Mexico and North America toward its summer home in Alaska. It first travels \(360 \mathrm{km}\) northwest to just off the coast of northern California, and then turns due north and travels \(400 \mathrm{km}\) toward its destination. Determine graphically the magnitude and direction of its displacement.

The sum of two vectors, \(\vec{A}+\vec{B},\) is perpendicular to their difference, \(\vec{A}-\vec{B}\). How do the vectors' magnitudes compare?

You throw a baseball at a \(45^{\circ}\) angle to the horizontal, aiming at a friend who's sitting in a tree a distance \(h\) above level ground. At the instant you throw your ball, your friend drops another ball. (a) Show that the two balls will collide, no matter what your ball's initial speed, provided it's greater than some minimum value, (b) Find an expression for that minimum speed,
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