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Heather in her Corvette accelerates at the rate of \((3.00 \hat{\mathrm{i}}-2.00 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}^{2},\) while Jill in her Jaguar accelerates at \((1.00 \hat{\mathrm{i}}+3.00 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}^{2} .\) 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 \((\mathrm{c})\) what is Heather's acceleration relative to Jill?

Short Answer

Expert verified
(a) Heather's speed with respect to Jill is approximately 26.93 m/s. (b) The distance between Heather and Jill after 5.00 s is approximately 134.54 m. (c) Heather's acceleration relative to Jill is \(2.00\hat{\mathrm{i}} - 5.00\hat{\mathrm{j}}\mathrm{m/s^2}\).

Step by step solution

01

Calculate Heather's velocity after 5 seconds

Heather's acceleration is given as \(3.00\hat{\mathrm{i}} - 2.00\hat{\mathrm{j}})\mathrm{m/s^2}\). Since she starts from rest, her velocity after \(t\) seconds can be found by \(\mathbf{v} = \mathbf{a}t\). After \(5.00\mathrm{s}\), her velocity \(\mathbf{v}_H\) is \(\mathbf{v}_H = (3.00\hat{\mathrm{i}} - 2.00\hat{\mathrm{j}})\mathrm{m/s^2} \times 5.00\mathrm{s} = (15\hat{\mathrm{i}} - 10\hat{\mathrm{j}})\mathrm{m/s}\).
02

Calculate Jill's velocity after 5 seconds

Jill's acceleration is given as \(1.00\hat{\mathrm{i}} + 3.00\hat{\mathrm{j}})\mathrm{m/s^2}\). Her velocity after \(5.00\mathrm{s}\) \(\mathbf{v}_J\) is \(\mathbf{v}_J = (1.00\hat{\mathrm{i}} + 3.00\hat{\mathrm{j}})\mathrm{m/s^2} \times 5.00\mathrm{s} = (5\hat{\mathrm{i}} + 15\hat{\mathrm{j}})\mathrm{m/s}\).
03

Find Heather's speed with respect to Jill

Heather's speed relative to Jill is the difference between their velocities: \((\mathbf{v}_H - \mathbf{v}_J) = (15\hat{\mathrm{i}} - 10\hat{\mathrm{j}}) - (5\hat{\mathrm{i}} + 15\hat{\mathrm{j}}) = (10\hat{\mathrm{i}} - 25\hat{\mathrm{j}})\mathrm{m/s}\). The magnitude of this relative velocity \(\mathbf{v}_{HJ}\) gives Heather's speed with respect to Jill: \(v_{HJ} = \sqrt{(10)^2 + (-25)^2} = \sqrt{100 + 625} = \sqrt{725}\mathrm{m/s}\).
04

Calculate the distance between Heather and Jill after 5 seconds

Heather's displacement after 5 seconds is \(\mathbf{d}_H = \mathbf{v}_H \times 5.00s = (15\hat{\mathrm{i}} - 10\hat{\mathrm{j}})\times 5 = (75\hat{\mathrm{i}} - 50\hat{\mathrm{j}})\mathrm{m}\). Jill's displacement is \(\mathbf{d}_J = \mathbf{v}_J \times 5.00s = (5\hat{\mathrm{i}} + 15\hat{\mathrm{j}})\times 5 = (25\hat{\mathrm{i}} + 75\hat{\mathrm{j}}\mathrm{m}\). The relative displacement is \((\mathbf{d}_H - \mathbf{d}_J) = (75\hat{\mathrm{i}} - 50\hat{\mathrm{j}}) - (25\hat{\mathrm{i}} + 75\hat{\mathrm{j}}) = (50\hat{\mathrm{i}} - 125\hat{\mathrm{j}})\mathrm{m}\). The distance between them is the magnitude of the relative displacement: \(|\mathbf{d}_{HJ}| = \sqrt{(50)^2 + (-125)^2} = \sqrt{2500 + 15625} = \sqrt{18125}\mathrm{m}\).
05

Determine Heather's acceleration relative to Jill

Heather's acceleration relative to Jill is the difference between their accelerations: \((\mathbf{a}_H - \mathbf{a}_J) = (3.00\hat{\mathrm{i}} - 2.00\hat{\mathrm{j}}) - (1.00\hat{\mathrm{i}} + 3.00\hat{\mathrm{j}}) = (2.00\hat{\mathrm{i}} - 5.00\hat{\mathrm{j}})\mathrm{m/s^2}\). So Heather's acceleration relative to Jill is \(2.00\hat{\mathrm{i}} - 5.00\hat{\mathrm{j}}\mathrm{m/s^2}\).

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Key Concepts

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

Understanding Kinematics in Physics
Kinematics is the branch of classical mechanics that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without consideration of the causes of motion. It involves analyzing velocity, acceleration, displacement, and time.

In the exercise given, we explore the motion of two cars, one driven by Heather and the other by Jill. As we are focused on kinematics, we look at how their accelerations affect their velocities over time. Since the cars are starting from rest, the final velocity is simply the product of acceleration and time, described by the equation \(\mathbf{v} = \mathbf{a}t\). Kinematics allows us to predict the future position and velocity of an object traveling with constant acceleration, which is exactly what we see in Heather and Jill's scenario.
Vector Subtraction in Relative Motion
Vector subtraction is a crucial concept in physics when dealing with quantities that have both a magnitude and a direction. It allows us to find the difference between two vectors which in turn gives us quantities like relative velocity or relative displacement.

In the context of the exercise, Heather and Jill's accelerations and velocities are given in the form of vectors, denoted by units \(\hat{\mathrm{i}}\) and \(\hat{\mathrm{j}}\) corresponding to the x and y components, respectively. When we calculate Heather's speed with respect to Jill, we use vector subtraction to determine their relative velocity. Similarly, vector subtraction is used to find the relative displacement between the two after a certain period, giving us insight into how far apart they are, despite starting from the same point.
Acceleration and Its Effects
Acceleration is the rate at which an object changes its velocity. It is a vector, which means it has both magnitude and direction. It is an essential concept in understanding motion, since even a constant acceleration can dramatically change an object's velocity over time.

In our given problem, Heather and Jill are both accelerating in different directions, shown by their acceleration vectors. To find Heather's acceleration relative to Jill, we subtract Jill's acceleration vector from Heather's. This relative acceleration shows how much quicker Heather's speed is changing compared to Jill's. The importance of acceleration cannot be understated, as it does not only change the speed of an object, but can also change its direction of motion if the acceleration is not aligned with the velocity.

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