To find the acceleration of a glider moving down a sloping air track, you measure its velocity at two points (v1 and v2); and the time t it takes between them: 3. v1 = 0.21 ± 0.05 m/s v2 = 0.85 + 0.05 m/2 t = 8.0 ± 0.1 s a. Assuming all uncertainties are independent and random, and acceleration is calculated using a = vz – Vi what should you report for a and its uncertainty?
To find the acceleration of a glider moving down a sloping air track, you measure its velocity at two points (v1 and v2); and the time t it takes between them: 3. v1 = 0.21 ± 0.05 m/s v2 = 0.85 + 0.05 m/2 t = 8.0 ± 0.1 s a. Assuming all uncertainties are independent and random, and acceleration is calculated using a = vz – Vi what should you report for a and its uncertainty?
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3. Please help me answer parts A. and B. of this question.
![### Finding the Acceleration of a Glider on a Sloping Air Track
To determine the acceleration of a glider moving down a sloping air track, you measure its velocity at two points (\(v_1\) and \(v_2\)), and the time (\(t\)) it takes between these points:
\[
v_1 = 0.21 \pm 0.05 \, \text{m/s}
\]
\[
v_2 = 0.85 \pm 0.05 \, \text{m/s}
\]
\[
t = 8.0 \pm 0.1 \, \text{s}
\]
#### Task:
a. Assuming all uncertainties are independent and random, and acceleration is calculated using \( a = \frac{v_2 - v_1}{t} \), what should you report for \(a\) and its uncertainty?
b. Based on an air resistance model, the predicted acceleration should be \( 0.13 \pm 0.01 \, \text{m/s}^2 \). Does your measurement agree with this prediction?
#### Explanation:
To calculate acceleration \(a\):
\[
a = \frac{v_2 - v_1}{t}
\]
Given values:
\[
v_1 = 0.21 \pm 0.05 \, \text{m/s}
\]
\[
v_2 = 0.85 \pm 0.05 \, \text{m/s}
\]
\[
t = 8.0 \pm 0.1 \, \text{s}
\]
Calculate the central value of \(a\):
\[
a = \frac{0.85 \, \text{m/s} - 0.21 \, \text{m/s}}{8.0 \, \text{s}} = \frac{0.64 \, \text{m/s}}{8.0 \, \text{s}} = 0.08 \, \text{m/s}^2
\]
To find the uncertainty in \(a\), we use the formula for the propagation of uncertainties for division:
\[
\left(\frac{\Delta a}{a}\right)^2 = \left(\frac{\Delta (v_2 - v_1)}{v_2 - v_1}\right)^2 + \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb3bae496-b29b-443e-8307-d9fa541b6fb0%2F47e059fc-bb85-4ead-a9f0-00f06e5db8dd%2Fotv8q03_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Finding the Acceleration of a Glider on a Sloping Air Track
To determine the acceleration of a glider moving down a sloping air track, you measure its velocity at two points (\(v_1\) and \(v_2\)), and the time (\(t\)) it takes between these points:
\[
v_1 = 0.21 \pm 0.05 \, \text{m/s}
\]
\[
v_2 = 0.85 \pm 0.05 \, \text{m/s}
\]
\[
t = 8.0 \pm 0.1 \, \text{s}
\]
#### Task:
a. Assuming all uncertainties are independent and random, and acceleration is calculated using \( a = \frac{v_2 - v_1}{t} \), what should you report for \(a\) and its uncertainty?
b. Based on an air resistance model, the predicted acceleration should be \( 0.13 \pm 0.01 \, \text{m/s}^2 \). Does your measurement agree with this prediction?
#### Explanation:
To calculate acceleration \(a\):
\[
a = \frac{v_2 - v_1}{t}
\]
Given values:
\[
v_1 = 0.21 \pm 0.05 \, \text{m/s}
\]
\[
v_2 = 0.85 \pm 0.05 \, \text{m/s}
\]
\[
t = 8.0 \pm 0.1 \, \text{s}
\]
Calculate the central value of \(a\):
\[
a = \frac{0.85 \, \text{m/s} - 0.21 \, \text{m/s}}{8.0 \, \text{s}} = \frac{0.64 \, \text{m/s}}{8.0 \, \text{s}} = 0.08 \, \text{m/s}^2
\]
To find the uncertainty in \(a\), we use the formula for the propagation of uncertainties for division:
\[
\left(\frac{\Delta a}{a}\right)^2 = \left(\frac{\Delta (v_2 - v_1)}{v_2 - v_1}\right)^2 + \
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