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 0.37 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 2.1 mm. If the floor is carpeted, this stopping distance is increased to about 1.0 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.
To calculate the magnitude and duration of deceleration in both cases, we can use the equations of motion.
1. Calculation for the hardwood floor:
The initial velocity (u) of the child's head is 0 m/s as it rolls off the bed.
The stopping distance (s) is 2.1 mm, which is equal to 0.0021 m.
The final velocity (v) is 0 m/s as the child's head comes to rest.
Using the equation v^2 = u^2 + 2as, we can solve for the acceleration (a):
0^2 = 0^2 + 2 * a * 0.0021
0 = 0 + 0.0042a
a = 0 / 0.0042
a = 0 m/s^2
Since the acceleration is 0 m/s^2, the child's head does not experience any deceleration on the hardwood floor, indicating no risk of injury.
2. Calculation for the carpeted floor:
The initial velocity (u) and final velocity (v) remain the same as before (0 m/s).
The stopping distance (s) is now 1.0 cm, which is equal to 0.01 m.
Using the same equation, v^2 = u^2 + 2as, we can solve for the acceleration (a):
0^2 = 0^2 + 2 * a * 0.01
0 = 0 + 0.02a
a = 0 / 0.02
a = 0 m/s^2
Similar to the previous case, the child's head does not experience any deceleration on the carpeted floor, indicating no risk of injury.
In both cases, the magnitude and duration of deceleration are found to be 0 m/s^2. Therefore, based on the given information, there is no risk of injury to the child from rolling off the bed onto either the hardwood or carpeted floor.
To solve this problem, we'll use the kinematic equation that relates the stopping distance (d), initial velocity (v), and deceleration (a):
d = (v^2) / (2a)
We can rearrange this formula to solve for deceleration:
a = (v^2) / (2d)
In the given scenario, we are given the stopping distance for both the hardwood floor (d1 = 2.1 mm = 0.0021 m) and the carpeted floor (d2 = 1.0 cm = 0.01 m), and we need to find the deceleration (a) for each case.
Let's calculate the deceleration for the hardwood floor:
Using the formula, a = (v^2) / (2d), we need to find the initial velocity (v).
We can find the initial velocity using the equation:
v = √(2ad)
v = √(2 * 800 m/s^2 * 0.0021 m)
v ≈ √3.36 m^2/s^2
v ≈ 1.83 m/s
Now that we have the initial velocity, we can calculate the deceleration:
a1 = (v^2) / (2d1)
a1 = (1.83 m/s)^2 / (2 * 0.0021 m)
a1 ≈ 15680.67 m/s^2
Now let's calculate the deceleration for the carpeted floor:
Using the same formula, a = (v^2) / (2d), we need to find the initial velocity (v) again.
v = √(2ad)
v = √(2 * 800 m/s^2 * 0.01 m)
v ≈ √16 m^2/s^2
v ≈ 4 m/s
Now we can determine the deceleration:
a2 = (v^2) / (2d2)
a2 = (4 m/s)^2 / (2 * 0.01 m)
a2 ≈ 800 m/s^2
Therefore, for the child rolling off the bed, the deceleration is approximately 15680.67 m/s^2 on a hardwood floor and 800 m/s^2 on a carpeted floor.
To determine the risk of injury, we need to compare these values to the acceleration thresholds given in the problem. Since both the hardwood and carpeted floors have accelerations greater than 1000 m/s^2, lasting for at least 1 ms, there is a risk of injury in both cases.
Calculaqte the speed of the head hitting the floor using
V = sqrt (2gH) , where H = 0.37 m
Let the stopping distance be X.
The deceleration rate required to stop the fall is
a = V^2/(2X) = 2gH/2X = 2 g H/X
The duration of deceleration is
t = X/(V/2)= 2H/V
since V/2 is the average velcoty during deceleration.
Use those formulas to come up with the numbers you need for the two floor coverings.