In a head on collision,a car stops in 0.18s from a speed of 29 m/s . The driver has a mass of 74 kg, and is, fortunately, tightly strapped into his seat. What force is applied to the driver by his seat belt during that fraction of a second?

F = M*a

M is the Mass and a is the acceleration, which in this case is
29 /0.18 = 161 m/s^2

The answer (F) will be in Newtons.
One Newton is about 0.22 lb force.

To find the force applied to the driver by his seat belt during the collision, we can use Newton's second law of motion, which states that force (F) is equal to the mass (m) multiplied by the acceleration (a). In this case, the acceleration can be calculated using the formula:

a = (change in velocity) / (time taken)

First, we need to calculate the change in velocity, which we can find using the initial velocity (vi) and the final velocity (vf). Since the car stops, the final velocity will be 0 m/s. The initial velocity is given as 29 m/s. Therefore, the change in velocity is:

change in velocity (Δv) = vf - vi
= 0 - 29
= -29 m/s (negative because the car stops)

The next step is to calculate the acceleration by dividing the change in velocity by the time taken:

acceleration (a) = Δv / t
= (-29 m/s) / (0.18 s)
= -161 m/s^2

Note: The negative acceleration represents the deceleration or slowing down of the car.

Now that we have the acceleration, we can calculate the force applied to the driver by their seat belt using Newton's second law:

force (F) = mass (m) * acceleration (a)
= 74 kg * (-161 m/s^2)
≈ -11914 N (rounded to two decimal places)

The negative sign indicates that the force is acting in the opposite direction to the motion of the driver, providing a restraining force to prevent them from moving forward.