Question
A 6-column table with 1 row. The first column labeled Number of Washers has entry 2. The second column labeled initial velocity v subscript 1 (meters per second) has entry 0.13. The third column labeled final velocity v subscript 2 (meters per second) has entry 0.36. The fourth column labeled Time to travel 0.25 meters t subscript 1 (seconds) has entry 1.92. The fifth column labeled time to travel 0.5 meters t subscript 2 (seconds) has entry 2.61. The sixth column labeled Acceleration a = StartFraction (v subscript 2 minus v subscript 1) over (t subscript 2 minus t subscript 1) EndFraction (meters per second squared) has entry empty.
The acceleration of the car with the data in the table above would be
m/s2.
If the applied force were cut in half, what do you predict the acceleration would be?
The acceleration of the car with the data in the table above would be
m/s2.
If the applied force were cut in half, what do you predict the acceleration would be?
Answers
Answered by
GPT-4o mini
To find the acceleration \( a \) using the data provided, we can apply the formula:
\[
a = \frac{v_2 - v_1}{t_2 - t_1}
\]
where:
- \( v_1 \) is the initial velocity (0.13 m/s),
- \( v_2 \) is the final velocity (0.36 m/s),
- \( t_1 \) is the time to travel the first distance (1.92 seconds),
- \( t_2 \) is the time to travel the second distance (2.61 seconds).
First, let's find the change in velocity:
\[
v_2 - v_1 = 0.36 \, \text{m/s} - 0.13 \, \text{m/s} = 0.23 \, \text{m/s}
\]
Next, let's find the change in time:
\[
t_2 - t_1 = 2.61 \, \text{s} - 1.92 \, \text{s} = 0.69 \, \text{s}
\]
Now we can substitute these values into the acceleration formula:
\[
a = \frac{0.23 \, \text{m/s}}{0.69 \, \text{s}} \approx 0.3333 \, \text{m/s}^2
\]
Therefore, the acceleration of the car is approximately:
\[
a \approx 0.33 \, \text{m/s}^2
\]
---
Now, if the applied force is cut in half, we can use Newton's second law of motion:
\[
F = m \cdot a
\]
If the force is halved and the mass of the car stays the same, the acceleration would also be halved. Thus, the predicted new acceleration \( a' \) would be:
\[
a' = \frac{1}{2} a = \frac{1}{2} \cdot 0.3333 \approx 0.1667 \, \text{m/s}^2
\]
So, if the applied force were cut in half, the predicted acceleration would be approximately:
\[
a' \approx 0.17 \, \text{m/s}^2
\]
\[
a = \frac{v_2 - v_1}{t_2 - t_1}
\]
where:
- \( v_1 \) is the initial velocity (0.13 m/s),
- \( v_2 \) is the final velocity (0.36 m/s),
- \( t_1 \) is the time to travel the first distance (1.92 seconds),
- \( t_2 \) is the time to travel the second distance (2.61 seconds).
First, let's find the change in velocity:
\[
v_2 - v_1 = 0.36 \, \text{m/s} - 0.13 \, \text{m/s} = 0.23 \, \text{m/s}
\]
Next, let's find the change in time:
\[
t_2 - t_1 = 2.61 \, \text{s} - 1.92 \, \text{s} = 0.69 \, \text{s}
\]
Now we can substitute these values into the acceleration formula:
\[
a = \frac{0.23 \, \text{m/s}}{0.69 \, \text{s}} \approx 0.3333 \, \text{m/s}^2
\]
Therefore, the acceleration of the car is approximately:
\[
a \approx 0.33 \, \text{m/s}^2
\]
---
Now, if the applied force is cut in half, we can use Newton's second law of motion:
\[
F = m \cdot a
\]
If the force is halved and the mass of the car stays the same, the acceleration would also be halved. Thus, the predicted new acceleration \( a' \) would be:
\[
a' = \frac{1}{2} a = \frac{1}{2} \cdot 0.3333 \approx 0.1667 \, \text{m/s}^2
\]
So, if the applied force were cut in half, the predicted acceleration would be approximately:
\[
a' \approx 0.17 \, \text{m/s}^2
\]