a car is proceeding at a speed of 14.0m/s when it collides with a stationary car in front. during the collision the first car moves a distance of 0.300 m as it comes to a stop. the driver is wearing her seat belt, so she remains in her seat during the collision. if the drivers mass is 52kg, how much force does the belt exert on her during the collision? neglect any friction between the driver and the seat.

For this questions guys, replace the , with a decimal point.

To find the force exerted by the seat belt on the driver, we can use the equation:

Force (F) = Mass (m) × Acceleration (a)

First, we need to find the acceleration. We can use the equation of motion:

v² = u² + 2as

Where:
v = Final velocity = 0 m/s (since the car comes to a stop)
u = Initial velocity = 14.0 m/s
a = Acceleration
s = Distance = 0.300 m

Rearranging the equation to solve for acceleration:

a = (v² - u²) / (2s)
a = (0 - (14.0)²) / (2 × 0.300)
a = -196 / 0.600
a = -326.67 m/s²

Note: The negative sign indicates that the acceleration is in the opposite direction to the initial velocity.

Now that we have the acceleration, we can calculate the force exerted by the seat belt:

F = m × a
F = 52 kg × -326.67 m/s²
F ≈ -17,000 N

Therefore, the seat belt exerts a force of approximately 17,000 Newtons on the driver during the collision.

To calculate the force exerted by the seat belt on the driver during the collision, we can use Newton's second law of motion, which states that force (F) is equal to the product of mass (m) and acceleration (a), i.e., F = m * a.

In this case, the initial velocity of the car (v) is 14.0 m/s, and it comes to a stop after moving a distance of 0.300 m. We need to determine the acceleration experienced by the car.

From kinematic equations, we can use the formula:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s)
u = initial velocity (14.0 m/s)
a = acceleration
s = distance traveled (0.300 m)

Rearranging the equation to solve for acceleration (a), we have:

a = (v^2 - u^2) / (2s)

Substituting the given values, we get:

a = (0^2 - 14.0^2) / (2 * 0.300)

Simplifying, we find:

a = (-196.0) / 0.600
a = -326.67 m/s^2

The negative sign indicates that the acceleration is in the opposite direction of motion.

Now, we can calculate the force exerted by the seat belt using the formula F = m * a. Given that the driver's mass (m) is 52 kg:

F = 52 kg * (-326.67 m/s^2)
F = -17,000 N

The force exerted by the seat belt on the driver during the collision is approximately 17,000 Newtons (N) in the opposite direction of motion.

Note: The negative sign indicates that the force is opposite to the direction of the initial motion of the car.

Well, force is rate of change of momentum or m A if the mass is constant, so what is the change of momentum?

original momentum of driver = 52 *14
= 588 kg m/s
final momentum of driver = 0
so change = -588 kg m/s

How long did it take?
average speed during crash = 14/2 = 7 m/s
it went .3 meters
so .3 = 7 t
t = .3 m /7 m/s = .042857 seconds (wow, fast)

so F = rate of change of momentum = -588 kg m/s / .042857s = 13,720 Newtons