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Chapter 3: Problem 17

(I) The position of a particular particle as a function of time is given by $$ \vec{\mathbf{r}}=\left(9.60 t \hat{\mathbf{i}}+8.85 \hat{\mathbf{j}}-1.00 t^{2} \hat{\mathbf{k}}\right) \mathrm{m} $$ Determine the particles velocity and acceleration as a function of time.

Short Answer

Expert verified
Velocity: \( \vec{\mathbf{v}} = (9.60 \hat{\mathbf{i}} - 2.00t \hat{\mathbf{k}}) \text{ m/s} \); Acceleration: \( \vec{\mathbf{a}} = -2.00 \hat{\mathbf{k}} \text{ m/s}^2 \).

Step by step solution

01

Understand the Position Function

The position function is given as \( \vec{\mathbf{r}} = (9.60t \hat{\mathbf{i}} + 8.85 \hat{\mathbf{j}} - 1.00t^2 \hat{\mathbf{k}}) \text{ m} \), where \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}} \) are unit vectors in the x, y, and z directions respectively.
02

Find the Velocity Function

Velocity is the rate of change of position with respect to time. To find the velocity function, differentiate the position function \( \vec{\mathbf{r}} \) with respect to time \( t \).\[\vec{\mathbf{v}} = \frac{d\vec{\mathbf{r}}}{dt} = \frac{d}{dt}(9.60t \hat{\mathbf{i}} + 8.85 \hat{\mathbf{j}} - 1.00t^2 \hat{\mathbf{k}})\]Calculating the derivatives for each component:- \( \frac{d}{dt}(9.60t) = 9.60 \)- \( \frac{d}{dt}(8.85) = 0 \)- \( \frac{d}{dt}(-1.00t^2) = -2.00t \)Thus, the velocity function is: \[\vec{\mathbf{v}} = (9.60 \hat{\mathbf{i}} + 0 \hat{\mathbf{j}} - 2.00t \hat{\mathbf{k}}) \text{ m/s}\]
03

Find the Acceleration Function

Acceleration is the rate of change of velocity with respect to time. To find the acceleration function, differentiate the velocity function \( \vec{\mathbf{v}} \) with respect to time \( t \).\[\vec{\mathbf{a}} = \frac{d\vec{\mathbf{v}}}{dt} = \frac{d}{dt}(9.60 \hat{\mathbf{i}} + 0 \hat{\mathbf{j}} - 2.00t \hat{\mathbf{k}})\]Calculating the derivatives for each component:- \( \frac{d}{dt}(9.60) = 0 \)- \( \frac{d}{dt}(0) = 0 \)- \( \frac{d}{dt}(-2.00t) = -2.00 \)Thus, the acceleration function is: \[\vec{\mathbf{a}} = (0 \hat{\mathbf{i}} + 0 \hat{\mathbf{j}} - 2.00 \hat{\mathbf{k}}) \text{ m/s}^2\]

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Key Concepts

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

Particle Motion
When studying kinematics, understanding particle motion is essential. It refers to how a particle moves in space over time. For this problem, we observe a particle moving in three-dimensional space, indicated by the unit vectors \(\hat{\mathbf{i}}, \hat{\mathbf{j}},\) and \(\hat{\mathbf{k}}\). These vectors represent movement along the x, y, and z-axes, respectively. The motion of the particle is a function of time, meaning its position changes at different time points. By analyzing its position, velocity, and acceleration functions, we learn the dynamic behavior of the particle. Let's delve into these core elements.
Position Function
The position function, \( \vec{\mathbf{r}} = (9.60t \hat{\mathbf{i}} + 8.85 \hat{\mathbf{j}} - 1.00t^2 \hat{\mathbf{k}}) \), is a mathematical representation of the particle's location in space at any given time \( t \).
  • The term \(9.60t \hat{\mathbf{i}}\) indicates the x-component, showing that the particle's x-position increases linearly with time.
  • The constant term \(8.85 \hat{\mathbf{j}}\) suggests a fixed y-position, meaning the particle does not move along the y-axis.
  • Finally, the \(-1.00t^2 \hat{\mathbf{k}}\) component signifies a parabolic motion in the z-direction, as it contains \(t^2\), showing motion akin to free-falling objects.
The position function gives a complete map of the particle's journey at any instant.
Velocity Function
Velocity describes how fast the position changes with time鈥攊t's the derivative of the position function. For this particle, after differentiating:\[\vec{\mathbf{v}} = (9.60 \hat{\mathbf{i}} + 0 \hat{\mathbf{j}} - 2.00t \hat{\mathbf{k}}) \text{ m/s}\]
  • The constant \(9.60\) means the particle moves consistently at this speed along the x-axis.
  • The y-component remains zero, confirming that there's no motion along the y-axis.
  • The term \(-2.00t\) indicates the z-component slows down linearly, suggesting a deceleration in the z-direction.
Understanding the velocity function helps visualize how the particle's motion unfolds over time, adjusting its speed in various directions.
Acceleration Function
Acceleration indicates how quickly the velocity changes over time鈥攊t's the derivative of the velocity function. Here, the equation is simplified to: \[\vec{\mathbf{a}} = (0 \hat{\mathbf{i}} + 0 \hat{\mathbf{j}} - 2.00 \hat{\mathbf{k}}) \text{ m/s}^2\]
  • The zero \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) components imply no acceleration along the x or y directions.
  • The constant \(-2.00 \hat{\mathbf{k}}\) component shows uniform acceleration in the negative z-direction.
This steady z-acceleration can be likened to the force of gravity acting in free-fall scenarios. By understanding the acceleration function, we grasp the forces influencing the particle's movement.

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Most popular questions from this chapter

A car is moving with speed \(18.0 \mathrm{~m} / \mathrm{s}\) due south at one moment and \(27.5 \mathrm{~m} / \mathrm{s}\) due east \(8.00 \mathrm{~s}\) later. Over this time interval, determine the magnitude and direction of \((a)\) its average velocity, \((b)\) its average acceleration. \((c)\) What is its average speed. [Hint: Can you determine all these from the information given?]

Derive a formula for the horizontal range \(R,\) of a projectile when it lands at a height \(h\) above its initial point. (For \(h<0\), it lands a distance \(-h\) below the starting point.) Assume it is projected at an angle \(\theta_{0}\) with initial speed \(v_{0}\).

A ball is shot from the top of a building with an initial velocity of \(18 \mathrm{~m} / \mathrm{s}\) at an angle \(\theta=42^{\circ}\) above the horizontal. (a) What are the horizontal and vertical components of the initial velocity? \((b)\) If a nearby building is the same height and \(55 \mathrm{~m}\) away, how far below the top of the building will the ball strike the nearby building?

(II) A football is kicked at ground level with a spced of 18.0 \(\mathrm{m} / \mathrm{s}\) at an angle of \(38.0^{\circ}\) to the horizontal. How much later does it hit the ground?

(1I) An ant walks on a piece of graph paper straight along the \(x\) axis a distance of 10.0 \(\mathrm{cm}\) in 2.00 \(\mathrm{s}\) . It then turns left \(30.0^{\circ}\) and walks in a straight line another 10.0 \(\mathrm{cm}\) in 1.80 s Finally, it turns another \(70.0^{\circ}\) to the left and walks another 10.0 \(\mathrm{cm}\) in 1.55 \(\mathrm{s}\) . Determine \((a)\) the \(x\) and \(y\) components of the ant's average velocity, and \((b)\) its magnitude and direction.
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