Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.38 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.9 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume that the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate.....

Hardwood floor magnitude___ m/s2
duration___ ms
Carpeted floor magnitude___m/s2
duration___ms

To calculate the magnitude and duration of the deceleration in both cases, we can use the equations of motion.

Let's start with the case of the child falling onto a hardwood floor.

1. Magnitude of deceleration on a hardwood floor:
We can use the equation of motion: vf^2 = vi^2 + 2ad, where vf is the final velocity (0 m/s), vi is the initial velocity, a is the acceleration, and d is the distance.
Here, the initial velocity is calculated using the equation of motion: vi^2 = u^2 + 2as, where u is the initial velocity (0 m/s), s is the displacement (0.38 m), and a is the acceleration.
Rearranging the equation, we get: a = (vf^2 - vi^2) / (2d).
Since the final velocity (vf) is 0 m/s, we plug in the given values and calculate the magnitude of the deceleration: a = (0 - 0^2) / (2 * 0.38). This simplifies to a = 0 m/s^2.

2. Duration of deceleration on a hardwood floor:
We know that the stopping distance is approximately 1.9 mm, which is equivalent to 0.0019 m. The duration of deceleration can be calculated using the equation of motion: vf = vi + at, where vf is the final velocity (0 m/s), vi is the initial velocity, a is the acceleration, and t is the time.
Since vf is 0 m/s and a is 0 m/s^2, we can rearrange the equation to solve for t: t = (vf - vi) / a.
Plugging in the given values, we have t = (0 - vi) / 0.
Since the acceleration is 0 m/s^2, the time taken to decelerate is infinite, meaning there is no deceleration on the hardwood floor.

Now let's move on to the case of the child falling onto a carpeted floor.

1. Magnitude of deceleration on a carpeted floor:
Using the same equation of motion as before: a = (vf^2 - vi^2) / (2d), we can plug in the values.
The stopping distance is approximately 1.1 cm, which is equivalent to 0.011 m.
Since the final velocity (vf) is 0 m/s, we have: a = (0 - 0^2) / (2 * 0.011). This simplifies to a = 0 m/s^2.

2. Duration of deceleration on a carpeted floor:
Using the equation of motion: t = (vf - vi) / a, we can calculate the time taken to decelerate.
Since vf is 0 m/s and a is 0 m/s^2, the time taken to decelerate is infinite, meaning there is no deceleration on the carpeted floor either.

In both cases, the magnitude of the deceleration and the duration of deceleration are both 0 m/s^2 and infinite, respectively. Therefore, there is no risk of injury in either case. This suggests that falling from a bed at a height of 0.38 m onto either a hardwood or carpeted floor will not cause a traumatic brain injury such as a concussion, according to the given acceleration thresholds.