/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 A web page designer creates an a... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 3

A web page designer creates an animation in which a dot on a computer screen has a position of \(\vec{r}=[4.0 \mathrm{cm}+\) \(\left(2.5 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2} ] \hat{\imath}+(5.0 \mathrm{cm} / \mathrm{s}) \hat{t}\) . (a) Find the magnitude and direction of the dot's average velocity between \(t=0\) and \(t=2.0 \mathrm{s} .\) (b) Find the magnitude and direction of the instantaneous velocity at \(t=0, t=1.0 \mathrm{s},\) and \(t=2.0 \mathrm{s}\) . (c) Sketch the dot's trajectory from \(t=0\) to \(t=2.0 \mathrm{s}\) , and show the velocities calculated in part \((\mathrm{b})\) .

Short Answer

Expert verified
The average velocity is 7.07 cm/s at 45掳. Instantaneous velocities are at 5.0 cm/s, 45掳 for \(t=1\) s, and 11.2 cm/s, 26.6掳 for \(t=2\) s.

Step by step solution

01

Analyze the Position Function

The position vector \( \vec{r} \) is given as a function of time: \( \vec{r} = [4.0 \text{ cm} + (2.5 \text{ cm/s}^2) t^2] \hat{\imath} + (5.0 \text{ cm/s}) t \hat{\jmath} \). This position function describes motion in the \(x\) and \(y\) directions.
02

Calculate Average Velocity

The average velocity between two times \(t_1\) and \(t_2\) is given by the change in position over the change in time: \[ \vec{v}_{\text{avg}} = \frac{\vec{r}(t_2) - \vec{r}(t_1)}{t_2 - t_1}. \] Compute \(\vec{r}(0)\) and \(\vec{r}(2)\) to find \( \vec{r}(0) = 4.0 \hat{\imath} \) and \( \vec{r}(2) = [4.0 + 10.0] \hat{\imath} + 10.0 \hat{\jmath} = 14.0 \hat{\imath} + 10.0 \hat{\jmath} \). Thus, \[ \vec{v}_{\text{avg}} = \frac{(14.0 \hat{\imath} + 10.0 \hat{\jmath}) - 4.0 \hat{\imath}}{2} = 5.0 \hat{\imath} + 5.0 \hat{\jmath} \text{ cm/s}. \]
03

Find Magnitude and Direction of Average Velocity

The magnitude of \( \vec{v}_{\text{avg}} \) is given by \( \| \vec{v}_{\text{avg}} \| = \sqrt{(5.0)^2 + (5.0)^2} = \sqrt{50} = 7.07 \text{ cm/s} \). The direction \( \theta \) can be found using \( \theta = \tan^{-1}\left( \frac{5.0}{5.0} \right) = 45^\circ \).
04

Calculate Instantaneous Velocity

Instantaneous velocity \( \vec{v}(t) \) is the derivative of \( \vec{r}(t) \): \[ \vec{v}(t) = \frac{d}{dt} \left( [4.0 + 2.5t^2] \hat{\imath} + 5.0t \hat{\jmath} \right) = (5.0t) \hat{\imath} + 5.0 \hat{\jmath}. \]
05

Calculate Instantaneous Velocities at Specific Times

- At \(t = 0\) s, \(\vec{v}(0) = 0 \hat{\imath} + 5.0 \hat{\jmath}\).- At \(t = 1.0\) s, \(\vec{v}(1) = 5.0 \hat{\imath} + 5.0 \hat{\jmath}\).- At \(t = 2.0\) s, \(\vec{v}(2) = 10.0 \hat{\imath} + 5.0 \hat{\jmath}\).Calculate magnitudes and directions similarly to previous steps.
06

Sketch the Dot鈥檚 Trajectory

Plot the position of the dot at \(t=0, 1, 2\) s using its position vectors and indicate instantaneous velocities as vectors originating from these points. The trajectory should indicate an upward curve due to the motion in both \(x\) and \(y\) directions.

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Key Concepts

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

Average Velocity
Average velocity refers to the overall change in position of an object divided by the total time taken. It's a vector quantity, which means it includes both magnitude and direction. For the problem at hand, this involves calculating how the position of the dot changes on a computer screen between given times.
Consider the position function of a moving object, which is given by the vector equation \( \vec{r}(t) = [4.0 \, \text{cm} + (2.5 \, \text{cm/s}^2) t^2] \hat{\imath} + 5.0 \, \text{cm/s} \cdot t \hat{\jmath} \).
  • To find the average velocity between times \( t_1 \) and \( t_2 \), use the formula\[ \vec{v}_{\text{avg}} = \frac{\vec{r}(t_2) - \vec{r}(t_1)}{t_2 - t_1} \].
  • Substituting in the values for \( t = 0 \) and \( t = 2.0 \, \text{s} \), we find the corresponding position vectors and ultimately the average velocity to be\( 5.0 \hat{\imath} + 5.0 \hat{\jmath} \, \text{cm/s} \).
  • Lastly, to determine the magnitude of this vector, apply the Pythagorean theorem:\[ \| \vec{v}_{\text{avg}} \| = \sqrt{(5.0)^2 + (5.0)^2} = 7.07 \, \text{cm/s} \].
  • The direction is determined by the angle \( \theta = \tan^{-1}\left( \frac{5.0}{5.0} \right) \), resulting in a 45-degree angle from the horizontal.
Instantaneous Velocity
Instantaneous velocity provides a snapshot of an object's velocity at a specific instant of time. Unlike average velocity, it involves taking the derivative of the position function with respect to time.
Given the position function \( \vec{r}(t) = [4.0 \, \text{cm} + (2.5 \, \text{cm/s}^2) t^2] \hat{\imath} + 5.0 \, \text{cm/s} \cdot t \hat{\jmath} \),
we find instantaneous velocity by differentiating with respect to time:
  • The instantaneous velocity function is \[ \vec{v}(t) = \frac{d}{dt} [4.0 + 2.5t^2] \hat{\imath} + \frac{d}{dt} (5.0t) \hat{\jmath} = (5.0t) \hat{\imath} + 5.0 \hat{\jmath} \].
  • Evaluate this at different times:
    • At \( t = 0 \) s, \( \vec{v}(0) = 0 \hat{\imath} + 5.0 \hat{\jmath} \).
    • At \( t = 1.0 \) s, \( \vec{v}(1) = 5.0 \hat{\imath} + 5.0 \hat{\jmath} \).
    • At \( t = 2.0 \) s, \( \vec{v}(2) = 10.0 \hat{\imath} + 5.0 \hat{\jmath} \).
  • Each vector's magnitude and direction can be calculated similarly to the average velocity.
Understanding instantaneous velocity is crucial when analyzing how an object's motion changes over small periods.
Trajectory
In physics, a trajectory describes the path that a moving object follows. For our problem, the dot's trajectory on the screen is determined by its position function over time. The trajectory offers valuable insight into the nature of the motion, indicating both direction and displacement.
To sketch the trajectory of the dot:
  • Identify the position of the dot at key intervals (e.g., \( t=0 \), \( t=1 \), \( t=2\) seconds) using its position vector function \( \vec{r}(t) \).
  • At each time point, graphically represent the instantaneous velocities, which show the speed and direction of the dot at each moment.
  • The movement trajectory appears as an upward bend in the plane since there is both horizontal and vertical motion.
  • This path gives a complete picture of how the dot moves across the screen from start to finish.
Visualizing trajectories helps in comprehending not just how fast or where an object travels, but also how its direction might dynamically evolve throughout its journey.

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Most popular questions from this chapter

A shot putter releases the shot some distance above the level ground with a velocity of \(12.0 \mathrm{m} / \mathrm{s}, 51.0^{\circ}\) above the horizontal. The shot hits the ground 2.08 \(\mathrm{s}\) later. You can ignore air resistance. (a) What are the components of the shot's acceleration while in flight? (b) What are the components of the shot's velocity at the beginning and at the end of its trajectory? (c) How far did she throw the shot horizontally? (d) Why does the expression for \(R\) in Example 3.8 not give the correct answer for part (c)? (e) How high was the shot above the ground when she released it? Draw \(x-t\) , \(y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the motion.

A rock tied to a rope moves in the \(x y\) -plane. Its coordinates are given as functions of time by $$ x(t)=R \cos \omega t \quad y(t)=R \sin \omega t $$ where \(R\) and \(\omega\) are constants. (a) Show that the rock's distance from the origin is constant and equal to \(R-\) that is, that its path is a circle of radius \(R .(b)\) Show that at every point the rock's velocity is perpendicular to its position vector. (c) Show that the rock's acceleration is always opposite in direction to its position vector and has magnitude \(\omega^{2} R .\) (d) Show that the magnitude of the rock's velocity is constant and equal to \(\omega R\) . (e) Combine the results of parts \((\mathrm{c})\) and \((\mathrm{d})\) to show that the rock's acceleration has constant magnitude \(v^{2} / R .\)

On the Flying Trapeze. new circus act is called the Texas Tumblers. Lovely Mary Belle swings from a trapeze, projects herself at an angle of \(53^{\circ},\) and is supposed to be canght by Joe Bob, whose hands are 6.1 \(\mathrm{m}\) above and 8.2 \(\mathrm{m}\) horizontally from her launch point (Fig. 3.50 ) You can ignore air resistance. (a) What initial speed \(v_{0}\) must Mary Belle have just to reach Joe Bob? (b) For the initial speed calculated in part (a), what are the magnitude and direction of her velocity when Mary Belle reaches Joe Bob? (c) Assuming that Mary Belle has the initial speed calculated in part (a), draw \(x-t, y-t, v_{x}-t,\) and \(v_{y}-t\) graphs showing the motion of both tumblers. Your graphs should show the motion up until the point where Mary Belle reaches Joe Bob. (d) The night of their debut performance, Joe Bob misses her completely as she flies past. How far horizontally does Mary Belle travel, from her initial launch point, before landing in the safety net 8.6 \(\mathrm{m}\) below her starting point?

A cannon, located 60.0 \(\mathrm{m}\) from the base of a vertical 25.0 -m- tall cliff, shoots a \(15-\mathrm{kg}\) shell at \(43.0^{\circ}\) above the horizontal toward the cliff. (a) What must the minimum muzzle velocity be for the shell to clear the top of the cliff? (b) The ground at the top of the cliff is level, with a constant elevation of 25.0 \(\mathrm{m}\) above the cannon. Under the conditions of part (a), how far does the shell land past the edge of the cliff?

A railroad flatear is traveling to the right at a speed of 13.0 \(\mathrm{m} / \mathrm{s}\) relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar (Fig. 3.43\()\) . What is the velocity (magnitude and direction) of the motor scooter relative to the flatcar if its velocity relative to the observer on the ground is (a) 18.0 \(\mathrm{m} / \mathrm{s}\) to the right? (b) 3.0 \(\mathrm{m} / \mathrm{s}\) to the left? (c) zero?
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