A table is slanted so a book resting on that table begins to accelerate at a constant value, \(a\).
If the book travels 45 cm in 1.0 seconds, determine its acceleration, \(a\)? (2 marks)
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A table is slanted so a book resting on that table begins to accelerate at a constant value, \(a\).
If the book travels 45 cm in 1.0 seconds, determine its acceleration, \(a\)? (2 marks)
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\(0.9\ \text{ms}^{-2}\ \ \text{down slope}\)
\(u= 0\ \text{ms}^{-1}, \ t=1\ \text{s}, \ s = \dfrac{45}{100} = 0.45\ \text{m}\)
| \(s\) | \(=ut +\dfrac{1}{2}at^2\) | |
| \(0.45\) | \(=0 + 0.5 \times a \times 1^2\) | |
| \(a\) | \(=\dfrac{0.45}{0.5 \times 1^2}\) | |
| \(=0.9\ \text{ms}^{-2}\ \ \text{down slope}\) |
Calculate the average acceleration of an airplane during landing if it touches down with a velocity of 60 m/s north and comes to a complete stop over a distance of 350 m. (2 marks)
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\(5.14\ \text{ms}^{-2}\) to the south.
| \(v^2\) | \(=u^2 +2as\) | |
| \(a\) | \(=\dfrac{v^2-u^2}{2s}\) | |
| \(=\dfrac{0-60^2}{2 \times 350}\) | ||
| \(=-5.14\ \text{ms}^{-2}\) |
Outline an experimental procedure to determine the acceleration of a falling steel ball. Your explanation should include all the measurements that must be recorded, the calculations needed to compute the acceleration, and an identification of any potential sources of error in the experiment. (6 marks)
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Procedure and Measurements:
Sources of Error:
Procedure and Measurements:
Sources of Error:
Using the velocity-time graph below
Part 1
Determine the magnitude of the displacement:
Part 2
Determine the average acceleration between 4 and 8 seconds:
Part 1: \(C\)
Part 2: \(B\)
Part 1
\(\text{Area}\ =(4 \times 8) + (\dfrac{1}{2} \times 4 \times 8) = 32 +16 = 48\ \text{m}\)
\(\Rightarrow C\)
Part 2
\(a= \dfrac{\Delta v}{\Delta t} = \dfrac{0-8}{8-4} = -2\ \text{ms}^{-2}\)
\(\Rightarrow B\)
An airplane, initially moving at 15 m/s, accelerates for 12 seconds until it reaches a take-off speed of 75 m/s. What is its average acceleration?
\(C\)
Average acceleration:
\(\dfrac{\Delta v}{\Delta t} = \dfrac{75-15}{12} = 5\ \text{ms}^{-2}\)
\(\Rightarrow C\)
A plane is travelling at 315 ms\(^{-1}\) north when it passes through a dense cloud and slows down to a velocity of 265 ms\(^{-1}\) for safety precautions.
The plane did not change direction and travelled 2.5 km while it was slowing down.
Using north as the positive direction for all calculations, determine:
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a. \(\text{50 ms}^{-1}\ \text{south}\)
b. \(\text{5.8 ms}^{-2}\ \text{south}\)
c. \(\text{8.62 s}\)
| a. | \(\Delta v\) | \(=v-u\) |
| \(=265-315\) | ||
| \(=-50\ \text{ms}^{-1}\) | ||
| \(=50\ \text{ms}^{-1}\ \text{south}\) |
b. Using \(v^2=u^2 +2as\) (time is not given):
| \(a\) | \(=\dfrac{v^2-u^2}{2s}\) | |
| \(=\dfrac{(265)^2-(315)^2}{2 \times 2500}\) | ||
| \(=-5.8\ \text{ms}^{-2}\) | ||
| \(=5.8\ \text{ms}^{-2}\) to the south. |
c. Using \(v=u+at\):
| \(t\) | \(=\dfrac{v-u}{a}\) | |
| \(=\dfrac{265-315}{-5.8}\) | ||
| \(=8.62\ \text{s}\) |
A truck travelling in a straight line with a speed of 60 ms\(^{-1}\) slows down and comes to rest over a period of 20 seconds.
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a. \(-60\ \text{ms}^{-1}\)
b. \(-3\ \text{ms}^{-2}\)
| a. | \(\Delta v\) | \(=v-u\) |
| \(=0-60\) | ||
| \(=-60\ \text{ms}^{-1}\) |
| b. | \(a\) | \(=\dfrac{v-u}{t}\) |
| \(=\dfrac{-60}{20}\) | ||
| \(=-3\ \text{ms}^{-2}\) |
This set of data was obtained from a motion investigation to determine the acceleration due to gravity on a planet other than Earth. --- 5 WORK AREA LINES (style=lined) ---
Time (s)
Vertical velocity (m s\(^{-1}\))
0.60
0.02
1.00
0.09
1.20
0.12
1.40
0.17
1.80
0.23
Plot the data from the table, and then calculate the acceleration. (3 marks)
--- 8 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- a. Timing of a falling mass. b. Assessing accuracy of results: c. Increasing data reliability: d. Assessing data reliability: a. Timing of a falling mass. b. Assessing accuracy of results: c. Increasing data reliability: d. Assessing data reliability: