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Heather in her Corvette accelerates at the rate of \((3.00 \hat{\mathbf{i}}-2.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2},\) while Jill in her Jaguar accelerates at \((1.00 \hat{\mathbf{i}}+3.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2} .\) They both start from rest at the origin of an \(x y\) coordinate system. After \(5.00 \mathrm{s},\) (a) what is Heather's speed with respect to Jill, (b) how far apart are they, and (c) what is Heather's acceleration relative to Jill?

Short Answer

Expert verified
(a) The relative speed of Heather with respect to Jill after 5 seconds is \(26.9 \mathrm{m/s}\). (b) The distance between them after 5 seconds is \(67.7 \mathrm{m}\). (c) The relative acceleration of Heather with respect to Jill is \(\left(2.00 \hat{\mathbf{i}} - 5.00 \hat{\mathbf{j}}\right) \mathrm{m/s^2}\).

Step by step solution

01

Determine Heather's and Jill's velocity vectors

Since both Heather and Jill are starting from rest, their initial velocity vectors are \(0 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}\). The velocity after time \(t\) for constant acceleration is calculated as initial velocity + acceleration*time. Since acceleration is constant, Heather's velocity at \(5 \mathrm{s}\) is \( \left(3.00 \hat{\mathbf{i}}-2.00 \hat{\mathbf{j}}\right) * 5.00 \mathrm{s} = \left(15.0 \hat{\mathbf{i}} -10.0 \hat{\mathbf{j}}\right) \mathrm{m/s}\) and Jill's velocity at \(5 \mathrm{s}\) is \( \left(1.00 \hat{\mathbf{i}}+3.00 \hat{\mathbf{j}}\right) * 5.00 \mathrm{s} = \left(5.0 \hat{\mathbf{i}} +15.0 \hat{\mathbf{j}}\right) \mathrm{m/s}\)
02

Calculate Heather's speed with respect to Jill

The relative velocity of Heather with respect to Jill is given by the vector difference of their velocities, which results in \( \left(15.0 - 5.0\right) \hat{\mathbf{i}} + \left(-10.0 -15.0\right) \hat{\mathbf{j}} = \left(10.0 \hat{\mathbf{i}} -25.0 \hat{\mathbf{j}}\right) \mathrm{m/s}\). The speed is the magnitude of the velocity vector, calculated as the square root of the sum of the squares of its components. Therefore, the relative speed of Heather with respect to Jill after \(5 \mathrm{s}\) is \(\sqrt{10^2 + (-25)^2} = 26.9 \mathrm{m/s}\)
03

Calculate how far apart they are

The position after time \(t\) under constant acceleration is given by \(0.5*a*t^2\), taking into account that both started from rest. Therefore, Heather's and Jill's positions after \(5 \mathrm{s}\) are \(\left(0.5 * 3.00^2*5^2\right) \hat{\mathbf{i}}- \left(0.5 * 2.00^2*5^2\right) \hat{\mathbf{j}} = \left(37.5 \hat{\mathbf{i}}- 25.0 \hat{\mathbf{j}}\right) \mathrm{m}\) and \(\left(0.5 * 1.00^2*5^2\right) \hat{\mathbf{i}}+ \left(0.5 * 3.00^2*5^2\right) \hat{\mathbf{j}} = \left(12.5 \hat{\mathbf{i}} + 37.5 \hat{\mathbf{j}}\right) \mathrm{m}\), respectively. The displacement between them is again the vector difference of their positions, yielding \(\left(37.5 - 12.5\right) \hat{\mathbf{i}} + \left(-25.0 - 37.5\right) \hat{\mathbf{j}} = \left(25.0 \hat{\mathbf{i}} -62.5 \hat{\mathbf{j}}\right) \mathrm{m}\). The magnitude of this vector gives the distance between them which is \(\sqrt{25^2 + (-62.5)^2} = 67.7 \mathrm{m}\)
04

Calculate Heather's acceleration relative to Jill

The relative acceleration of Heather with respect to Jill is simply the vector difference of their accelerations, which is \(\left(3.00 - 1.00\right) \hat{\mathbf{i}} + \left(-2.00 - 3.00\right) \hat{\mathbf{j}} = \left(2.00 \hat{\mathbf{i}} - 5.00 \hat{\mathbf{j}}\right) \mathrm{m/s^2}\)

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Key Concepts

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

Acceleration
Acceleration is a concept often encountered in physics, representing the rate of change of velocity over time. In simple terms, it tells us how quickly an object is speeding up or slowing down. The unit for acceleration is meters per second squared \( ext{m/s}^2 \). Heather's Corvette and Jill's Jaguar each have given accelerations in this exercise. Heather's acceleration vector is given as \((3.00 \hat{\mathbf{i}} - 2.00 \hat{\mathbf{j}}) \mathrm{m/s^2}\) and Jill's as \((1.00 \hat{\mathbf{i}} + 3.00 \hat{\mathbf{j}}) \mathrm{m/s^2}\).
The acceleration vector consists of two components: one in the x-direction (i) and one in the y-direction (j), which represent the acceleration in those directions on a 2D plane. This shows us not only how fast each car is accelerating but also the direction in which each component of the acceleration is acting.
Understanding acceleration vectors is crucial to solving problems involving motion, as they help differentiate between the speeds and directions of the objects in question.
Velocity
Velocity is another fundamental concept in the study of motion, defined as the speed of an object in a given direction. Unlike speed, which is scalar, velocity is a vector property, meaning it has both magnitude and direction.In the situation described in the exercise, both Heather and Jill start from rest. This implies that their initial velocities are both zero. Over time, due to their respective accelerations, they develop velocities.
The velocity of Heather after 5 seconds can be derived from her acceleration and time, resulting in \( (3.00 \hat{\mathbf{i}} - 2.00 \hat{\mathbf{j}}) \times 5.00 \text{ s} = (15.0 \hat{\mathbf{i}} - 10.0 \hat{\mathbf{j}}) \text{ m/s} \).Similarly, Jill's velocity becomes \( (1.00 \hat{\mathbf{i}} + 3.00 \hat{\mathbf{j}}) \times 5.00 \text{ s} = (5.0 \hat{\mathbf{i}} + 15.0 \hat{\mathbf{j}}) \text{ m/s} \).
The concept of relative velocity plays a key role here, allowing us to find Heather's speed with respect to Jill. This is calculated by finding the difference between their velocities, providing insight into how fast one is moving in comparison to the other.
Coordinate System
A coordinate system is a framework that allows us to describe the positions of points in space in a mathematical way. In this particular problem, a 2D Cartesian coordinate system (x, y) is used to manage the motion of Heather and Jill.
Starting at the origin (point 0,0), both cars are described by their positions on this grid as time progresses, traced by their respective accelerations and velocities.The positions after 5 seconds are obtained by using the formula \( s = 0.5 \times a \times t^2 \). Heather's coordinates thus become \( 37.5 \hat{\mathbf{i}} - 25.0 \hat{\mathbf{j}} \) and Jill's \( 12.5 \hat{\mathbf{i}} + 37.5 \hat{\mathbf{j}} \).
These positional vectors lay out the path taken by the vehicles in our specified coordinate system and enable us to calculate how far apart they are, showing the practical application of 2D coordinates in real-world scenarios.
Vector Analysis
Vector analysis is a mathematical tool used to understand quantities that have both magnitude and direction, offering a way to evaluate physical phenomena like motion. It's especially needed in this scenario where multiple dimensions are involved.
Vectors, like the ones describing Heather’s and Jill’s accelerations and velocities, are instrumental when performing calculations, such as finding relative speed or distance. The vector difference between Heather's and Jill's respective velocities gives the relative velocity vector \( (10.0 \hat{\mathbf{i}} - 25.0 \hat{\mathbf{j}}) \mathrm{m/s} \).This is pivotal in finding how fast Heather is moving compared to Jill.
Additionally, vector subtraction and the Pythagorean theorem help in determining the physical separation between the two vehicles, and ultimately their relative motions. Vector analysis simplifies complex equations into more digestible parts, providing clarity in the calculations of physical quantities.

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Most popular questions from this chapter

The small archerfish (length 20 to \(25 \mathrm{cm}\) ) lives in brackish waters of southeast Asia from India to the Philippines. This aptly named creature captures its prey by shooting a stream of water drops at an insect, either flying or at rest. The bug falls into the water and the fish gobbles it up. The archerfish has high accuracy at distances of \(1.2 \mathrm{m}\) to \(1.5 \mathrm{m}\) and it sometimes makes hits at distances up to 3.5 m. A groove in the roof of its mouth, along with a curled tongue, forms a tube that enables the fish to impart high velocity to the water in its mouth when it suddenly closes its gill flaps. Suppose the archerfish shoots at a target \(2.00 \mathrm{m}\) away, at an angle of \(30.0^{\circ}\) above the horizontal. With what velocity must the water stream be launched if it is not to drop more than \(3.00 \mathrm{cm}\) vertically on its path to the target?

A playground is on the flat roof of a city school, \(6.00 \mathrm{m}\) above the street below. The vertical wall of the building is \(7.00 \mathrm{m}\) high, to form a meter-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of \(53.0^{\circ}\) above the horizontal at a point 24.0 meters from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. (a) Find the speed at which the ball was launched. (b) Find the vertical distance by which the ball clears the wall. (c) Find the distance from the wall to the point on the roof where the ball lands.

A fireworks rocket explodes at height \(h,\) the peak of its vertical trajectory. It throws out burning fragments in all directions, but all at the same speed \(v .\) Pellets of solidified metal fall to the ground without air resistance. Find the smallest angle that the final velocity of an impacting fragment makes with the horizontal.

A motorist drives south at \(20.0 \mathrm{m} / \mathrm{s}\) for \(3.00 \mathrm{min}\) then turns west and travels at \(25.0 \mathrm{m} / \mathrm{s}\) for \(2.00 \mathrm{min},\) and finally travels northwest at \(30.0 \mathrm{m} / \mathrm{s}\) for 1.00 min. For this 6.00 -min trip, find (a) the total vector displacement, (b) the average speed, and (c) the average velocity. Let the positive \(x\) axis point east.

Note: Ignore air resistance in all problems and take \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) at the Earth's surface. To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of \(300 \mathrm{m} / \mathrm{s}\) at \(55.0^{\circ}\) above the horizontal. It explodes on the mountainside \(42.0 \mathrm{s}\) after firing. What are the \(x\) and \(y\) coordinates of the shell where it explodes, relative to its firing point?

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