A small coin, initially at rest, begins falling. If the clock starts when the coin begins to fall, what is the magnitude of the coin's displacement between t1 = 0.226 s and 12 = 0.903 s? Ay = m
A small coin, initially at rest, begins falling. If the clock starts when the coin begins to fall, what is the magnitude of the coin's displacement between t1 = 0.226 s and 12 = 0.903 s? Ay = m
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Problem Statement:**
A small coin, initially at rest, begins falling. If the clock starts when the coin begins to fall, what is the magnitude of the coin's displacement between \( t_1 = 0.226 \, \text{s} \) and \( t_2 = 0.903 \, \text{s} \)?
**Solution Section:**
To find the magnitude of the coin's displacement \( \Delta y \), we use the equation for displacement in uniformly accelerated motion:
\[ \Delta y = v_i t + \frac{1}{2} a t^2 \]
Since the coin starts from rest, the initial velocity \( v_i = 0 \). The acceleration \( a \) is due to gravity, which is approximately \( 9.81 \, \text{m/s}^2 \).
The time interval \( \Delta t = t_2 - t_1 = 0.903 \, \text{s} - 0.226 \, \text{s} = 0.677 \, \text{s} \).
Thus, the displacement can be calculated by substituting these values into the formula.
\[ \Delta y = 0 + \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times (0.677 \, \text{s})^2 \]
Calculate the displacement to find \( \Delta y \).
**Answer Box:**
\[ \Delta y = \text{[Calculated Value]} \, \text{m} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F52caa5b4-58c6-470e-a25b-6f2dba4e23c8%2F29f22455-b8b5-415a-815b-e66548776522%2F76ht2b_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A small coin, initially at rest, begins falling. If the clock starts when the coin begins to fall, what is the magnitude of the coin's displacement between \( t_1 = 0.226 \, \text{s} \) and \( t_2 = 0.903 \, \text{s} \)?
**Solution Section:**
To find the magnitude of the coin's displacement \( \Delta y \), we use the equation for displacement in uniformly accelerated motion:
\[ \Delta y = v_i t + \frac{1}{2} a t^2 \]
Since the coin starts from rest, the initial velocity \( v_i = 0 \). The acceleration \( a \) is due to gravity, which is approximately \( 9.81 \, \text{m/s}^2 \).
The time interval \( \Delta t = t_2 - t_1 = 0.903 \, \text{s} - 0.226 \, \text{s} = 0.677 \, \text{s} \).
Thus, the displacement can be calculated by substituting these values into the formula.
\[ \Delta y = 0 + \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times (0.677 \, \text{s})^2 \]
Calculate the displacement to find \( \Delta y \).
**Answer Box:**
\[ \Delta y = \text{[Calculated Value]} \, \text{m} \]
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