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Chapter 4: Problem 13

A particle moves so that its position (in meters) as a function of time (in seconds) is \(\vec{r}=\hat{\mathrm{i}}+4 t^{2} \hat{\mathrm{j}}+t \hat{\mathrm{k}} .\) Write expressions for (a) its velocity and (b) its acceleration as functions of time.

Short Answer

Expert verified
Velocity: \(\vec{v}(t) = 8t\hat{\mathrm{j}} + \hat{\mathrm{k}}\); Acceleration: \(\vec{a}(t) = 8\hat{\mathrm{j}}\).

Step by step solution

01

Analyze Position Vector

The position vector of the particle is given as \(\vec{r}= \hat{\mathrm{i}} + 4t^2 \hat{\mathrm{j}} + t \hat{\mathrm{k}}\). Each component describes how the particle's position changes over time in the respective directions (\(\hat{\mathrm{i}}\), \(\hat{\mathrm{j}}\), and \(\hat{\mathrm{k}}\)).
02

Derive Velocity Vector

The velocity vector \(\vec{v}(t)\) is the first derivative of the position vector with respect to time. Differentiate each component separately: \(\frac{d}{dt}(\hat{\mathrm{i}}) = 0 \hat{\mathrm{i}}, \frac{d}{dt}(4t^2 \hat{\mathrm{j}}) = 8t \hat{\mathrm{j}}, \text{and } \frac{d}{dt}(t \hat{\mathrm{k}}) = 1 \hat{\mathrm{k}}\). Thus, \(\vec{v}(t) = 0\hat{\mathrm{i}} + 8t\hat{\mathrm{j}} + 1\hat{\mathrm{k}}\).
03

Derive Acceleration Vector

The acceleration vector \(\vec{a}(t)\) is the derivative of the velocity vector with respect to time. Differentiate the velocity components: \(\frac{d}{dt}(0 \hat{\mathrm{i}}) = 0 \hat{\mathrm{i}}, \frac{d}{dt}(8t \hat{\mathrm{j}}) = 8 \hat{\mathrm{j}}, \text{and } \frac{d}{dt}(1 \hat{\mathrm{k}}) = 0 \hat{\mathrm{k}}\). Therefore, \(\vec{a}(t) = 0\hat{\mathrm{i}} + 8\hat{\mathrm{j}} + 0\hat{\mathrm{k}}\).

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Key Concepts

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

Position Vector
The position vector is a fundamental concept in kinematics, representing the position of a particle in space as a function of time. In our exercise, the position vector is given as \( \vec{r} = \hat{\mathrm{i}} + 4t^2 \hat{\mathrm{j}} + t \hat{\mathrm{k}} \). This vector describes how the particle's position changes over time along three orthogonal directions: x-axis (\( \hat{\mathrm{i}} \)), y-axis (\( \hat{\mathrm{j}} \)), and z-axis (\( \hat{\mathrm{k}} \)).
The components of the vector indicate that:
  • The x-position is constant, \( \hat{\mathrm{i}} \), meaning there is no movement along the x-axis regardless of time.
  • The y-position changes as \( 4t^2 \hat{\mathrm{j}} \), indicating that the particle moves along the y-axis with a quadratic dependence on time. This suggests acceleration in the y direction as time progresses.
  • For the z-direction, the position component \( t \hat{\mathrm{k}} \) means that the particle moves linearly along the z-axis over time.
Understanding the position vector allows us to visualize and predict the path of the moving particle.
Velocity Vector
The velocity vector is crucial for understanding how quickly and in which direction the particle's position changes over time. It is the first derivative of the position vector with respect to time.
In our example, to determine the velocity vector \( \vec{v}(t) \), we differentiate \( \vec{r} \) and find:
  • The derivative of the x-component is zero, reflecting no change or movement in the x-direction.
  • The y-component changes as \( 8t \hat{\mathrm{j}} \), indicating that the velocity along the y-axis varies linearly with time. This result suggests the presence of acceleration along the y-axis.
  • The z-component's derivative is \( 1 \hat{\mathrm{k}} \), signifying a constant velocity in the z-direction.
The resulting velocity vector is \( \vec{v}(t) = 0\hat{\mathrm{i}} + 8t\hat{\mathrm{j}} + 1\hat{\mathrm{k}} \), showing that while there is no velocity component in the x-direction, the particle accelerates in the y-direction and maintains a steady velocity in the z-direction.
Acceleration Vector
The acceleration vector reveals how the velocity of the particle changes over time and is obtained by differentiating the velocity vector. It's a key factor in understanding the dynamic behavior of the particle.
For our particle:
  • The derivative of the velocity's x-component is zero, meaning there's no acceleration in the x-direction.
  • The y-component of the velocity differentiates to \( 8 \hat{\mathrm{j}} \), which indicates a constant acceleration in the y-direction. The value 8 is the rate at which the velocity along the y-axis increases per second.
  • The z-component's derivative is zero, showing no change in velocity, hence no acceleration in the z-direction.
Thus, the acceleration vector is \( \vec{a}(t) = 0\hat{\mathrm{i}} + 8\hat{\mathrm{j}} + 0\hat{\mathrm{k}} \). This distinct acceleration in the y-direction influences the particle to speed up as it travels along the y-axis, shaping its overall trajectory in the plane.

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Most popular questions from this chapter

A football kicker can give the ball an initial speed of \(25 \mathrm{~m} / \mathrm{s} .\) What are the (a) least and (b) greatest elevation angles at which he can kick the ball to score a field goal from a point \(50 \mathrm{~m}\) in front of goalposts whose horizontal bar is \(3.44 \mathrm{~m}\) above the ground?

A projectile is fired with an initial speed \(v_{0}=30.0 \mathrm{~m} / \mathrm{s}\) from level ground at a target that is on the ground, at distance \(R=20.0 \mathrm{~m}\) as shown in Fig. \(4-59 .\) What are the (a) least and (b) greatest launch angles that will allow the projectile to hit the target?

The velocity \(\vec{v}\) of a particle moving in the \(x y\) plane is given by \(\vec{v}=\left(6.0 t-4.0 t^{2}\right) \hat{\mathrm{i}}+8.0 \hat{\mathrm{j}},\) with \(\vec{v}\) in meters per second and \(t(>0)\) in seconds. (a) What is the acceleration when \(t=3.0 \mathrm{~s} ?\) (b) When (if ever) is the acceleration zero? (c) When (if ever) is the velocity zero? (d) When (if ever) does the speed equal \(10 \mathrm{~m} / \mathrm{s} ?\)

A proton initially has \(\vec{v}=4.0 \hat{\mathrm{i}}-2.0 \hat{\mathrm{j}}+3.0 \hat{\mathrm{k}}\) and then \(4.0 \mathrm{~s}\) later has \(\vec{v}=-2.0 \hat{\mathrm{i}}-2.0 \hat{\mathrm{j}}+5.0 \hat{\mathrm{k}}\) (in meters per second). For that \(4.0 \mathrm{~s},\) what are \((\mathrm{a})\) the proton's average acceleration \(\vec{a}_{\mathrm{avg}}\) in unitvector notation, (b) the magnitude of \(\vec{a}_{\text {avg }},\) and (c) the angle between \(\vec{a}_{\text {avg }}\) and the positive direction of the \(x\) axis?

A car travels around a flat circle on the ground, at a constant speed of \(12.0 \mathrm{~m} / \mathrm{s}\). At a certain instant the car has an acceleration of \(3.00 \mathrm{~m} / \mathrm{s}^{2}\) toward the east. What are its distance and direction from the center of the circle at that instant if it is traveling (a) clockwise around the circle and (b) counterclockwise around the circle?
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