In a car crash, a 44 kilogram passenger moving at 15 meters per second is brought to rest by an air bag during a 0.10 second time interval. What is the magnitude of the average force exerted on the passenger during this time?
6600 N
Well, if we're talking about a car crash, I think it's safe to say that this is no time for clowning around! Safety is serious business. Now, to find the average force exerted on the passenger, we can use Newton's second law of motion: Force equals mass times acceleration. In this case, the acceleration is the change in velocity divided by the time interval. So, the magnitude of the average force exerted on the passenger during this time can be calculated by multiplying the mass of the passenger (44 kilograms) by the change in velocity (from 15 meters per second to zero) and then dividing it by the time interval (0.10 seconds). And the result would be... uh, hold on a second... math's not really my strong suit... let me call in my mathematician friend to help me out with this one. *Honks clown horn*
To find the magnitude of the average force exerted on the passenger, we can use Newton's second law of motion, which states that force is equal to mass times acceleration.
First, we need to calculate the acceleration of the passenger. Since the passenger is brought to rest, the final velocity is 0 m/s. Therefore, the change in velocity (Δv) can be calculated as:
Δv = final velocity - initial velocity
Δv = 0 m/s - 15 m/s
Δv = -15 m/s
Since the time interval (Δt) is given as 0.10 seconds, we can calculate the acceleration (a) using the formula:
a = Δv / Δt
a = (-15 m/s) / (0.10 s)
a = -150 m/s²
Now, we can calculate the magnitude of the average force (F) using the formula:
F = m * a
F = (44 kg) * (-150 m/s²)
F = - 6600 N
Therefore, the magnitude of the average force exerted on the passenger during this time is 6600 Newtons.
To find the magnitude of the average force exerted on the passenger, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration.
In this case, the passenger's initial velocity (v1) is 15 m/s and their final velocity (v2) is 0 m/s since they are brought to rest. The time taken (Δt) is 0.10 seconds.
First, we need to calculate the acceleration of the passenger using the formula:
acceleration (a) = (v2 - v1) / Δt
Substituting the given values:
a = (0 m/s - 15 m/s) / 0.10 s
a = -150 m/s²
This negative sign indicates that the acceleration is in the opposite direction of the initial velocity.
Next, we can calculate the force using the formula:
force (F) = mass (m) × acceleration (a)
Substituting the given mass of the passenger:
F = 44 kg × -150 m/s²
F = -6,600 N
The negative sign represents that the force exerted on the passenger is in the opposite direction of their initial motion. However, since we are looking for the magnitude of the force, we can ignore the negative sign.
Therefore, the magnitude of the average force exerted on the passenger during this time is 6,600 Newtons (N).
(average force)*(time interval) = (momentum change) = 44*15 = 660 kg m/s
Solve for the average force, which will be in newtons