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Chapter 4: Problem 89

A particle starts from the origin at \(t=0\) with a velocity of \(8.0 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}\) and moves in the \(x y\) plane with constant acceleration \((4.0 \hat{1}+2.0 \hat{j}) \mathrm{m} / \mathrm{s}^{2}\). When the particle's \(x\) coordinate is \(29 \mathrm{~m}\), what are its (a) \(y\) coordinate and (b) speed?

Short Answer

Expert verified
(a) The y-coordinate is approximately 52.34 m. (b) The speed is approximately 21.86 m/s.

Step by step solution

01

Write the equation for x-coordinate

Using the equation for motion with initial velocity and constant acceleration, the x-coordinate at any time \(t\) can be given by: \[ x = v_{0x}t + \frac{1}{2}a_x t^2 \]Since the initial velocity in the x-direction \(v_{0x}\) is 0 (as the velocity is solely in the \(j\) direction initially), the equation simplifies to:\[ x = \frac{1}{2} \times 4.0 \times t^2 \]Substitute \(x = 29\) m to find \(t\): \[ 29 = 2t^2 \] which gives us \(t^2 = 14.5\). Hence, \(t = \sqrt{14.5}\).
02

Calculate the value of t

Solving \(t = \sqrt{14.5}\) yields:\[ t \approx 3.807 \text{ s} \].
03

Write the equation for y-coordinate

Using the motion equation for the y-coordinate, we have:\[ y = v_{0y}t + \frac{1}{2}a_y t^2 \]Substitute \(v_{0y} = 8.0\) m/s and \(a_y = 2.0\) m/s虏 to get:\[ y = 8.0 \times t + \frac{1}{2} \times 2.0 \times t^2 = 8.0t + t^2 \]
04

Calculate the y-coordinate

Substitute \(t \approx 3.807\) s into the y-coordinate equation:\[ y = 8.0 \times 3.807 + (3.807)^2 \]Calculating this yields:\[ y \approx 52.336 \text{ m} \].
05

Write the equation for speed

The speed \(v\) can be found using the velocity components in the x and y directions. The x-component of velocity is \(v_{x} = a_x t\), and y-component of velocity is \(v_{y} = v_{0y} + a_y t\).Thus,\[ v_{x} = 4.0 \times t \]\[ v_{y} = 8.0 + 2.0 \times t \].
06

Calculate the speed

Substituting \(t \approx 3.807\) into the velocity components:\[ v_{x} = 4.0 \times 3.807 = 15.228 \text{ m/s} \]\[ v_{y} = 8.0 + 2.0 \times 3.807 = 15.614 \text{ m/s} \]The speed is found by combining these components:\[ v = \sqrt{(15.228)^2 + (15.614)^2} \approx 21.857 \text{ m/s} \].

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Key Concepts

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

Equation of motion
In 2D kinematics, the equation of motion is fundamental for predicting how a particle moves in a plane. This equation relates position, velocity, and acceleration over time. It is often expressed as:
  • In the x-direction: \[ x = v_{0x}t + \frac{1}{2}a_x t^2 \] This formula calculates the x-position of a particle after a time \( t \), starting from an initial velocity \( v_{0x} \) and experiencing constant acceleration \( a_x \).
  • In the y-direction: \[ y = v_{0y}t + \frac{1}{2}a_y t^2 \] Here, \( v_{0y} \) and \( a_y \) are the initial velocity and acceleration in the y-direction, respectively.
These equations allow us to determine the position at any moment in time. For example, substituting the given values (like when \( x = 29 \text{ m} \)) helps find specific points in the particle's trajectory, such as what time \( t \) the particle reaches that x-coordinate.
Velocity components
Velocity in 2D motion is decomposed into components along the x and y axes. Understanding these components is crucial as they are influenced by both initial velocity and acceleration. For a particle:
  • The x-component of velocity, \( v_{x} \), is influenced solely by acceleration in the x-direction if the initial velocity in that direction is zero: \[ v_{x} = a_x t \]
  • The y-component, \( v_{y} \), is affected by both the initial velocity and the acceleration: \[ v_{y} = v_{0y} + a_y t \]
By combining these components, the overall speed of the particle can be determined using the Pythagorean theorem: \[ v = \sqrt{v_{x}^2 + v_{y}^2} \]This relationship helps explain how the particle's motion is pieced together from each axis component, providing a full picture of its velocity at any given moment.
Constant acceleration
Constant acceleration simplifies the calculations involved in physics problems because it allows using straightforward equations. This consistency means that the equations of motion maintain a uniform structure regardless of how complex the motion might otherwise be.
  • With constant acceleration, the velocity changes linearly over time.
  • The equations of motion can be directly applied, avoiding any need for more complex calculus approaches.
  • For example, a constant acceleration in our problem is given as \( a = (4.0 \hat{i} + 2.0 \hat{j}) \text{ m/s}^2 \).
By using this constant value, determining the particle鈥檚 exact position or velocity at any time becomes a matter of simply substituting into a formula, keeping computations straightforward and predictable.
Physics problem solving
Solving physics problems, especially in kinematics, involves a strategic method to ensure all aspects of motion are considered.
  • Identifying known and unknown quantities: This involves listing what's given in the problem and what needs solving. For example, the initial velocities and accelerations in different directions were provided.
  • Applying appropriate equations: Once the knowns and unknowns are clear, use equations of motion or expressions for velocity to link the quantities.
  • Substituting and solving: Replacing the variables with the given numbers, like the x-coordinate or time, helps calculate other needed values accurately.
  • Rechecking calculations: Ensure accuracy by verifying each step, especially in substituting and simplifying expressions. This avoids errors and strengthens understanding.
This systematic approach helps break down complex 2D motion problems into manageable portions, ensuring a complete and correct solution.

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Most popular questions from this chapter

Suppose that a shot putter can put a shot at the world-class speed \(v_{0}=15.00 \mathrm{~m} / \mathrm{s}\) and at a height of \(2.160 \mathrm{~m} .\) What horizontal distance would the shot travel if the launch angle \(\theta_{0}\) is (a) \(45.00^{\circ}\) and (b) \(42.00^{\circ}\) ? The answers indicate that the angle of \(45^{\circ}\), which maximizes the range of projectile motion, does not maximize the horizontal distance when the launch and landing are at different heights.

A plane, diving with constant speed at an angle of \(53.0^{\circ}\) with the vertical, releases a projectile at an altitude of \(730 \mathrm{~m}\). The projectile hits the ground \(5.00 \mathrm{~s}\) after release. (a) What is the speed of the plane? (b) How far does the projectile travel horizontally during its flight? What are the (c) horizontal and (d) vertical components of its velocity just before striking the ground?

A dart is thrown horizontally with an initial speed of \(10 \mathrm{~m} / \mathrm{s}\) toward point \(P\), the bull's-eye on a dart board. It hits at point \(Q\) on the rim, vertically below \(P, 0.19 \mathrm{~s}\) later. (a) What is the distance \(P Q ?\) (b) How far away from the dart board is the dart released?

A woman who can row a boat at \(6.4 \mathrm{~km} / \mathrm{h}\) in still water faces a long, straight river with a width of \(6.4 \mathrm{~km}\) and a current of \(3.2 \mathrm{~km} / \mathrm{h}\). Let i point directly across the river and \(\hat{j}\) point directly downstream. If she rows in a straight line to a point directly opposite her starting position, (a) at what angle to i must she point the boat and (b) how long will she take? (c) How long will she take if, instead, she rows \(3.2 \mathrm{~km}\) down the river and then back to her starting point? (d) How long if she rows \(3.2 \mathrm{~km} u p\) the river and then back to her starting point? (e) At what angle to \(\hat{\mathrm{i}}\) should she point the boat if she wants to cross the river in the shortest possible time? (f) How long is that shortest time?

An Earth satellite moves in a circular orbit \(640 \mathrm{~km}\) above Earth's surface with a period of \(98.0 \mathrm{~min}\). What are the (a) speed and (b) magnitude of the centripetal acceleration of the satellite?
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