Question
A research team is examining the braking system of a newly designed car model. The 1950- kilogram car moves along a track at a speed of 22.0 meters per second. Force is subsequently applied to bring the car to a full stop. This is depicted in the provided data, along with the time required for the car to come to a halt.
Trial
Force Applied (N)
Time Required to Stop (s)
1
10,725
4.00
2
5,362.5
8.00
In what way does the information substantiate the equation provided in Newtop's sifcond law of motion, F=ma?
(1 point)
Acceleration will rise as the time needed for the stopping of motion increases
Applying double the force to the car will lead to its deceleration at a rate twice as fast
Applying double the force to the car will result in it decelerating at half the rate
The acceleration rate will remain constant for a car with consistent mass
Trial
Force Applied (N)
Time Required to Stop (s)
1
10,725
4.00
2
5,362.5
8.00
In what way does the information substantiate the equation provided in Newtop's sifcond law of motion, F=ma?
(1 point)
Acceleration will rise as the time needed for the stopping of motion increases
Applying double the force to the car will lead to its deceleration at a rate twice as fast
Applying double the force to the car will result in it decelerating at half the rate
The acceleration rate will remain constant for a car with consistent mass
Answers
GPT-4o mini
To analyze the information provided in relation to Newton's second law of motion, \( F = ma \), we will consider the relationship between the applied force, the mass of the car, and the resulting acceleration (or deceleration in this case).
Here's a breakdown of the trials:
1. **Trial 1:**
- Mass of the car, \( m = 1950 \) kg
- Force applied, \( F_1 = 10,725 \) N
- Time to stop, \( t_1 = 4.00 \) s
We can calculate the acceleration (using \( a = \frac{F}{m} \)):
\[
a_1 = \frac{F_1}{m} = \frac{10,725}{1950} \approx 5.49 \text{ m/s}^2
\]
We can also find the final velocity \( v_f = 0 \) and initial velocity \( v_i = 22.0 \) m/s.
Using the formula \( v_f = v_i + at \):
\[
0 = 22.0 + a_1 \cdot 4.0 \implies a_1 = -\frac{22.0}{4.0} = -5.5 \text{ m/s}^2
\]
2. **Trial 2:**
- Force applied, \( F_2 = 5,362.5 \) N
- Time to stop, \( t_2 = 8.00 \) s
Similar to trial 1:
\[
a_2 = \frac{F_2}{m} = \frac{5,362.5}{1950} \approx 2.75 \text{ m/s}^2
\]
Using the stopping condition:
\[
0 = 22.0 + a_2 \cdot 8.0 \implies a_2 = -\frac{22.0}{8.0} = -2.75 \text{ m/s}^2
\]
Now, we can relate the findings to the options regarding how they reflect Newton's second law \( F = ma \):
- **Option 1**: "Acceleration will rise as the time needed for the stopping of motion increases." - This is incorrect; as time increases, the car decelerates slowly (lower acceleration).
- **Option 2**: "Applying double the force to the car will lead to its deceleration at a rate twice as fast." - Correct, this aligns with Newton's second law, as doubling the force results in double the deceleration.
- **Option 3**: "Applying double the force to the car will result in it decelerating at half the rate." - This is incorrect, as increasing force leads to greater acceleration/deceleration, not half.
- **Option 4**: "The acceleration rate will remain constant for a car with consistent mass." - This is incorrect; the acceleration depends on the force applied, which varies.
The best matching statement regarding the findings is **Option 2**: Applying double the force to the car will lead to its deceleration at a rate twice as fast.
Here's a breakdown of the trials:
1. **Trial 1:**
- Mass of the car, \( m = 1950 \) kg
- Force applied, \( F_1 = 10,725 \) N
- Time to stop, \( t_1 = 4.00 \) s
We can calculate the acceleration (using \( a = \frac{F}{m} \)):
\[
a_1 = \frac{F_1}{m} = \frac{10,725}{1950} \approx 5.49 \text{ m/s}^2
\]
We can also find the final velocity \( v_f = 0 \) and initial velocity \( v_i = 22.0 \) m/s.
Using the formula \( v_f = v_i + at \):
\[
0 = 22.0 + a_1 \cdot 4.0 \implies a_1 = -\frac{22.0}{4.0} = -5.5 \text{ m/s}^2
\]
2. **Trial 2:**
- Force applied, \( F_2 = 5,362.5 \) N
- Time to stop, \( t_2 = 8.00 \) s
Similar to trial 1:
\[
a_2 = \frac{F_2}{m} = \frac{5,362.5}{1950} \approx 2.75 \text{ m/s}^2
\]
Using the stopping condition:
\[
0 = 22.0 + a_2 \cdot 8.0 \implies a_2 = -\frac{22.0}{8.0} = -2.75 \text{ m/s}^2
\]
Now, we can relate the findings to the options regarding how they reflect Newton's second law \( F = ma \):
- **Option 1**: "Acceleration will rise as the time needed for the stopping of motion increases." - This is incorrect; as time increases, the car decelerates slowly (lower acceleration).
- **Option 2**: "Applying double the force to the car will lead to its deceleration at a rate twice as fast." - Correct, this aligns with Newton's second law, as doubling the force results in double the deceleration.
- **Option 3**: "Applying double the force to the car will result in it decelerating at half the rate." - This is incorrect, as increasing force leads to greater acceleration/deceleration, not half.
- **Option 4**: "The acceleration rate will remain constant for a car with consistent mass." - This is incorrect; the acceleration depends on the force applied, which varies.
The best matching statement regarding the findings is **Option 2**: Applying double the force to the car will lead to its deceleration at a rate twice as fast.