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
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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:**

\[ \Delta y = \text{[Calculated Value]} \, \text{m} \]
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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