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.

Question

Trial force applied(N) Time required to stop(seconds)
1 10,725 4.00
2 5,362.5 8.00

In what way does the information substantiate the equation provided in Newton's second law of motion, F = ma?

(1 point)
Responses

Acceleration will rise as the time needed for the stopping of motion increases.
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 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.
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.
The acceleration rate will remain constant for a car with consistent mass.

Answer 1

To analyze the information provided and relate it to Newton's second law of motion (F = ma), let's examine the results from the trials:

1. Trial 1:

   • Force applied (F₁) = 10,725 N
   • Time to stop (t₁) = 4.00 seconds

2. Trial 2:

   • Force applied (F₂) = 5,362.5 N
   • Time to stop (t₂) = 8.00 seconds

Calculating Acceleration for Each Trial

Using the formula \( a = \frac{F}{m} \):

• For the first trial:

  • \( a₁ = \frac{10,725 , \text{N}}{1950 , \text{kg}} \approx 5.49 , \text{m/s}^2 \)

• For the second trial:

  • \( a₂ = \frac{5,362.5 , \text{N}}{1950 , \text{kg}} \approx 2.75 , \text{m/s}^2 \)

Understanding the Relationship

From the above calculations:

• The first trial applies a greater force than the second trial, which results in a greater acceleration in the opposite direction (deceleration in this context).
• When the force is doubled (as in moving from second trial to first trial), the acceleration (deceleration) also increases approximately twofold, confirming that the relationship between force and acceleration established by Newton's second law is upheld.

Selecting the Correct Response

Based on the analysis, the correct interpretation is:

Applying double the force to the car will lead to its deceleration at a rate twice as fast.

Conclusion

This option effectively indicates that as force increases, acceleration (or in this case, deceleration) increases proportionally, which is consistent with \( F = ma \).