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:
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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 \( v_f = v_i + at \): \[ 0 = 22.0 + a_1 \cdot 4.0 \implies a_1 = -\frac{22.0}{4.0} = -5.5 \text{ m/s}^2 \]
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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 \( F = ma \):
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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).
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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.
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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.
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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.