In an emergency stop to avoid an accident, a shoulder-strap seatbelt holds a 58 passenger in place. If the car was initially traveling at 94 and came to a stop in 5.2 along a straight, level road, what was the magnitude of the average force applied to the passenger by the seatbelt?
What was the direction of the average force applied to the passenger by the seatbelt?
To find the magnitude of the average force applied to the passenger by the seatbelt, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = m*a). In this case, the acceleration is the change in velocity divided by the time taken to stop.
Step 1: Calculate the acceleration of the car
The change in velocity can be calculated by subtracting the final velocity from the initial velocity: Δv = 0 - 94 = -94 m/s (the negative sign indicates a deceleration).
Next, divide the change in velocity by the time taken to stop: a = Δv / t = -94 / 5.2 = -18.08 m/s^2.
Step 2: Calculate the mass of the passenger
Given that the shoulder-strap seatbelt holds a 58 kg passenger in place, the mass of the passenger is 58 kg.
Step 3: Calculate the magnitude of the average force
Using Newton's second law (F = m*a), substitute the values we have:
F = (58 kg) * (-18.08 m/s^2) = -1049.44 N
The magnitude of the average force applied to the passenger is approximately 1049.44 N.
Now, let's determine the direction of the average force applied to the passenger. Since the direction of force is always in the opposite direction of acceleration (according to Newton's second law), the force applied to the passenger by the seatbelt is in the forward direction (opposite to the direction of the car's motion).