To understand how the provided data substantiates Newton's second law of motion, \( F = ma \), we need to analyze the situation using the concepts of force, mass, and acceleration.
Given Data:
- Mass of the car: \( m = 1950 \) kg
- Initial velocity of the car: \( v_i = 22.0 \) m/s
- Final velocity of the car (stop): \( v_f = 0 \) m/s
- Force applied and time taken to stop in two trials.
Calculating Deceleration:
Using the kinematic equation: \[ v_f = v_i + at \] We can rearrange to solve for acceleration \( a \) (which will be negative, as it is deceleration): \[ a = \frac{v_f - v_i}{t} = \frac{0 - 22.0 , \text{m/s}}{t} \]
Trial 1 Calculations:
- Force applied: \( F_1 = 10,725 \) N
- Time required to stop: \( t_1 = 4.00 \) s
- Acceleration: \[ a_1 = \frac{0 - 22.0 , \text{m/s}}{4.00 , \text{s}} = -5.5 , \text{m/s}^2 \]
- Using \( F = ma \): \[ F_1 = (1950 , \text{kg}) \times (-5.5 , \text{m/s}^2) \] \[ F_1 = -10,725 , \text{N} \] (The negative sign indicates the force is in the opposite direction of motion.)
Trial 2 Calculations:
- Force applied: \( F_2 = 5,362.5 \) N
- Time required to stop: \( t_2 = 8.00 \) s
- Acceleration: \[ a_2 = \frac{0 - 22.0 , \text{m/s}}{8.00 , \text{s}} = -2.75 , \text{m/s}^2 \]
- Using \( F = ma \): \[ F_2 = (1950 , \text{kg}) \times (-2.75 , \text{m/s}^2) \] \[ F_2 = -5,362.5 , \text{N} \]
Conclusion:
Both trials show that the applied force is consistent with Newton's second law of motion (\( F = ma \)). In each case, when we calculate the acceleration based on the time to stop and relate it back to the force using the car's mass, we find that the calculated force corresponds exactly to the force that was applied in both trials.
Thus, the information substantiates the equation \( F = ma \) by demonstrating that the forces applied to the car result in the observed deceleration, complying with the expected relationships dictated by Newton's laws.
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