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A bird is initially flying horizontally east at \(21.1 \mathrm{~m} / \mathrm{s}\), but one second later it has changed direction so that it is flying horizontally and \(7^{\circ}\) north of east, at the same speed. What are the magnitude and direction of its acceleration vector during that one second time interval? (Assume its acceleration was roughly constant.) (answer check available at lightandmatter.com)

Short Answer

Expert verified
The acceleration magnitude is approximately 2.57 m/s² at an angle of 3.5° north of east.

Step by step solution

01

Understand Initial and Final Velocities

Initially, the bird's velocity is entirely in the east direction at 21.1 m/s. After 1 second, its velocity is at a 7° angle north of east, but the magnitude remains 21.1 m/s. We need to find the components of these velocities.
02

Calculate Initial Velocity Components

The initial velocity is purely eastward, so its components are: \( v_{ix} = 21.1 \, \text{m/s} \) and \( v_{iy} = 0 \, \text{m/s} \).
03

Calculate Final Velocity Components

Using trigonometry for the final velocity vectors: \( v_{fx} = 21.1 \times \cos(7^{\circ}) \, \text{m/s} \) and \( v_{fy} = 21.1 \times \sin(7^{\circ}) \, \text{m/s} \). Calculate these values to get the components.
04

Calculate Change in Velocity

Find the changes in velocity components: \( \Delta v_x = v_{fx} - v_{ix} \) and \( \Delta v_y = v_{fy} - v_{iy} \).
05

Calculate Acceleration Components

Using \( a = \frac{\Delta v}{\Delta t} \) where \( \Delta t = 1 \, \text{s} \), calculate: \( a_x = \frac{\Delta v_x}{1} \) and \( a_y = \frac{\Delta v_y}{1} \).
06

Find Magnitude of Acceleration

Use Pythagorean theorem to find the magnitude: \( a = \sqrt{a_x^2 + a_y^2} \).
07

Find Direction of Acceleration

To find the direction: \( \theta = \arctan\left(\frac{a_y}{a_x}\right) \). This gives the angle north of east.

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Key Concepts

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

Velocity Components
When dealing with motion in physics, particularly in two dimensions, understanding the concept of velocity components is crucial. Velocity is a vector, meaning it has both a magnitude and a direction. To analyze its effects thoroughly, we often break it down into components – typically along the x (horizontal) and y (vertical) axes. In our exercise, the bird initially has a velocity of 21.1 m/s entirely in the east, which means all of its velocity is along the horizontal axis.
After a second, although the magnitude of the velocity remains the same, the direction changes by 7° north of east. To find out how this affects the motion, we decompose this velocity into two components:
  • Horizontal (eastward) component: \(v_{fx} = 21.1 \cos(7^\circ) \)
  • Vertical (northward) component: \(v_{fy} = 21.1 \sin(7^\circ) \)
These components tell us how much of the bird's velocity is along each axis, assisting in further calculations like acceleration.
Acceleration Calculation
Acceleration is all about how quickly an object's velocity changes over time. In this problem, since we assume constant acceleration, the bird’s change in velocity over the one-second interval directly relates to acceleration.
After calculating the change in velocity components, the acceleration components can be calculated as follows:
  • Horizontal acceleration \(a_x = \frac{\Delta v_x}{\Delta t}\)
  • Vertical acceleration \(a_y = \frac{\Delta v_y}{\Delta t}\)
Given \(\Delta t = 1\, \text{s}\), the changes in velocity components are simply equivalent to the acceleration components. Thus, the change from initial to final velocity dictates how the bird accelerates in both horizontal and vertical directions.
Kinematics
Kinematics is the study of motion without considering its causes. It focuses on the description of motion, which here includes the bird's velocity and acceleration.
In this scenario, the motion involves the initial and final velocities and how these change over time due to acceleration. By using kinematic equations, we can clearly determine the relationships between distance, velocity, acceleration, and time. The primary kinematic equation used here is:
  • \(\Delta v = a \cdot \Delta t\)
Understanding this equation allows us to connect the concepts of velocity change and acceleration, providing a complete picture of how the bird's flight path changes in the given time period.
Trigonometry in Physics
Trigonometry plays an essential role in physics, especially in resolving vector components. By using trigonometric functions such as sine and cosine, we can break down a vector into perpendicular components that are easier to work with.
In this exercise, the angle information provided (7° north of east) was critical in calculating the final velocity components. Here's how trigonometry helped:
  • To find the horizontal component \(v_{fx} = 21.1 \cos(7^\circ)\)
  • To find the vertical component \(v_{fy} = 21.1 \sin(7^\circ)\)
Using these trigonometric identities, we converted directional velocity into components that could be analyzed algebraically, aiding in the subsequent calculation of acceleration. Without trigonometry, resolving such vectors would be highly complex.

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Most popular questions from this chapter

A flexible rope of mass \(m\) and length \(L\) slides without friction over the edge of a table. Let \(x\) be the length of the rope that is hanging over the edge at a given moment in time. (a) Show that \(x\) satisfies the equation of motion \(d^{2} x / d t^{2}=g x / L .\) [Hint: Use \(F=d p / d t\), which allows you to handle the two parts of the rope separately even though mass is moving out of one part and into the other.] (b) Give a physical explanation for the fact that a larger value of \(x\) on the right-hand side of the equation leads to a greater value of the acceleration on the left side. (c) When we take the second derivative of the function \(x(t)\) we are supposed to get essentially the same function back again, except for a constant out in front. The function \(e^{x}\) has the property that it is unchanged by differentiation, so it is reasonable to look for solutions to this problem that are of the form \(x=b e^{c t}\), where \(b\) and \(c\) are constants. Show that this does indeed provide a solution for two specific values of \(c\) (and for any value of \(b\) ). (d) Show that the sum of any two solutions to the equation of motion is also a solution. (e) Find the solution for the case where the rope starts at rest at \(t=0\) with some nonzero value of \(x\).

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