2 part question, only looking for part b.
a) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.2 m. What is the net force acting on a 62 kg driver who is driving at 18 m/sec and comes to rest in this distance?
b) A drive who does not wear a selt belt continues to move at the initial velocity until she or he hits something solid (e.g the stearing wheel) and then comes to rest in a very short distance. Find the net force on a driver without seatbelts who comes to rest in 0.9 cm.
Use the following equation for both:
(Force)(Stopping distance) = (1/2) M V^2
V is the car velocity before collision.
I assume they want you to use the same M and V for part b)
To calculate the net force acting on the driver without a seatbelt in part b, we can use the same formula as in part a:
Force = mass * acceleration
First, let's find the acceleration. We know that the driver comes to rest in a distance of 0.9 cm. However, it is important to convert this distance to meters for consistency. So, 0.9 cm is equivalent to 0.009 m.
We can use the kinematic equation:
vf^2 = vi^2 + 2ad
where vf is the final velocity (0 m/s since the driver comes to rest), vi is the initial velocity (which is given as 18 m/s), a is the acceleration, and d is the distance.
Rearranging the equation to solve for acceleration (a):
a = (vf^2 - vi^2) / (2d)
Plugging in the values:
a = (0 - (18 m/s)^2) / (2 * 0.009 m)
Simplifying the equation:
a = (-324 m^2/s^2) / (0.018 m) = -18,000 m/s^2
Now we can calculate the net force:
Force = mass * acceleration
Plugging in the given mass of 62 kg and the calculated acceleration of -18,000 m/s^2:
Force = 62 kg * (-18,000 m/s^2)
Calculating the result:
Force = -1,116,000 N
So, the net force acting on the driver without seatbelts who comes to rest in a distance of 0.9 cm is approximately -1,116,000 N.