Final answer:
The average force required to stop a 1100-kg car in 8.0s is calculated using Newton's second law of motion. By assuming a change in velocity equivalent to the initial velocity over the stopping time, the force is determined to be 137.5 N (Option A).
Step-by-step explanation:
To find out the average force required to stop a 1100-kg car in 8.0 seconds, we need to use the formula derived from Newton's second law of motion, F = ma, where F is the force applied, m is the mass of the object, and a is the acceleration.
Firstly, we need to calculate the acceleration needed to bring the car to a stop. Assuming the car comes to rest from an initial velocity (v0), the final velocity (v) will be 0 since the car stops. The formula for acceleration when initial and final velocities are known is a = (v - v0) / t, where t is the time taken to stop.
However, we are not provided with the initial velocity of the car. But we know that the car comes to a complete stop, so we can infer that the change in velocity is equal to the initial velocity. Therefore, we can assume an initial velocity (v0) on the understanding that this quantity will cancel out.
Now, using the simplified formula, which doesn't require the initial velocity:
a = -v0 / t
Assuming that the car is moving forward and comes to a stop, the acceleration must be negative, meaning the car is decelerating.
Then we use F = ma to find the force:
F = m(-v0 / t)
Multiply the mass (m = 1100 kg) by the negative acceleration. We can remove the negative sign as we are looking for the magnitude of the force, not its direction.
So without the negative sign, the force would be:
F = 1100 kg * v0 / 8.0 s
This simplifies to multiplying the car's mass by the initial velocity and then dividing by the time:
F = (1100 kg * v0) / 8.0 s
The initial velocity (v0) is effectively 1 (since it's not given, we assume this to simplify and to highlight that it will cancel out), so:
F = (1100 kg * 1) / 8.0 s
Therefore, F = 137.5 N.
So, the correct answer is option A) 137.5 N.