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

1 answer

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.