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 (s)
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?
A. Applying double the force to the car will result in it decelerating at half the rate.
B. The acceleration rate will remain constant for a car with consistent mass.
C. Acceleration will rise as the time needed for the stopping of motion increases.
D. Applying double the force to the car will lead to its deceleration at a rate twice as fast.

Answer 1

To analyze the relationship between the force applied to the car, its mass, and its resulting acceleration, we can use Newton's second law of motion, which states that \( F = ma \), where \( F \) is the force exerted on an object, \( m \) is the mass of the object, and \( a \) is the acceleration of the object.

Given:

• Mass of the car, \( m = 1950 , \text{kg} \)
• Initial speed of the car, \( v = 22.0 , \text{m/s} \)
• The equations of motion reveal that acceleration can be calculated using the formula:

\[ a = \frac{F}{m} \]

From the trials:

1. Trial 1:

   • Force, \( F = 10725 , \text{N} \)
   • Time to stop, \( t = 4.00 , \text{s} \)

   The car's deceleration (\( a_1 \)): \[ a_1 = \frac{F}{m} = \frac{10725 , \text{N}}{1950 , \text{kg}} \approx 5.49 , \text{m/s}^2 \]

2. Trial 2:

   • Force, \( F = 5362.5 , \text{N} \)
   • Time to stop, \( t = 8.00 , \text{s} \)

   The car's deceleration (\( a_2 \)): \[ a_2 = \frac{F}{m} = \frac{5362.5 , \text{N}}{1950 , \text{kg}} \approx 2.75 , \text{m/s}^2 \]

Now, we can evaluate the given options regarding how they align with \( F = ma \):

A. Applying double the force to the car will result in it decelerating at half the rate.

• This statement is incorrect, as doubling the force would double the acceleration (deceleration in this case) not halve it.

B. The acceleration rate will remain constant for a car with consistent mass.

• This is misleading because while mass remains constant, the acceleration depends on the applied force. It can vary with different applied forces.

C. Acceleration will rise as the time needed for the stopping of motion increases.

• This is incorrect, as time and acceleration do not have a direct correlation in this sense without considering initial conditions and forces.

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

• This statement is correct since, according to Newton's second law, doubling the force should indeed double the acceleration (deceleration).

The best answer is therefore:

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