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 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 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.
Hardfloor Mag:
Hardfloor Duration:
Carpet Floor Mag:
Carpetfloor Duration:
U didn’t even answer the question
To calculate the magnitude and duration of deceleration in both cases, we can use the equation of motion:
v^2 = u^2 + 2as,
where v is the final velocity (in this case, zero), u is the initial velocity, a is the acceleration, and s is the displacement.
Let's first calculate the initial velocity (u) for both cases:
For the child falling from the bed onto the hardwood floor:
Initial velocity (u) = sqrt(2gh),
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the height of the bed (0.38 m).
u = sqrt(2 * 9.8 * 0.38) = 2.92 m/s (rounded to two decimal places)
For the child falling from the bed onto the carpeted floor:
The initial velocity will be the same, which is 2.92 m/s.
Now, let's calculate the magnitude and duration of deceleration for each case:
1. Hardwood Floor:
Stopping distance (s) = 2.1 mm = 0.0021 m
Using the equation v^2 = u^2 + 2as, we can solve for the deceleration (a):
0 = (2.92)^2 + 2a(0.0021)
Solving this equation for a gives us:
a = -((2.92)^2) / (2 * 0.0021) ≈ -1964.38 m/s^2 (rounded to two decimal places)
The negative sign indicates deceleration.
Since we know the initial velocity (u) and final velocity (v) are zero, we can use the equation v = u + at to find the duration (t) of deceleration:
0 = 2.92 - 1964.38 * t
Solving this equation for t gives us:
t = 2.92 / 1964.38 ≈ 0.00149 s (rounded to five decimal places)
Therefore, on the hardwood floor, the magnitude of the deceleration is approximately 1964.38 m/s^2, and it lasts for approximately 0.00149 seconds.
2. Carpeted Floor:
Stopping distance (s) = 1.0 cm = 0.01 m
Using the same equation, v^2 = u^2 + 2as, we can solve for the deceleration (a):
0 = (2.92)^2 + 2a(0.01)
Solving this equation for a gives us:
a = -((2.92)^2) / (2 * 0.01) ≈ -425.44 m/s^2 (rounded to two decimal places)
Similarly, v = u + at can be used to find the duration (t) of deceleration:
0 = 2.92 - 425.44 * t
Solving this equation for t gives us:
t = 2.92 / 425.44 ≈ 0.006861 s (rounded to six decimal places)
Therefore, on the carpeted floor, the magnitude of the deceleration is approximately 425.44 m/s^2, and it lasts for approximately 0.006861 seconds.
To determine the risk of injury based on the given guidelines, we can compare the magnitudes and durations of the decelerations to the threshold criteria:
For a traumatic brain injury to occur:
- The acceleration must be greater than 1000 m/s^2.
AND
- The duration must be at least 1 ms (0.001 s).
Comparing the cases:
1. Hardwood Floor:
Magnitude: 1964.38 m/s^2
Duration: 0.00149 s
2. Carpeted Floor:
Magnitude: 425.44 m/s^2
Duration: 0.006861 s
In both cases, the magnitude of the deceleration is greater than the threshold of 1000 m/s^2. However, only the deceleration on the hardwood floor meets the duration requirement of 1 ms.
Therefore, based on the given guidelines, there is a risk of injury when the child falls from the bed onto both the hardwood floor and the carpeted floor. However, the risk might be relatively higher on the hardwood floor due to the shorter duration of deceleration.
find velocity when hitting the floor.
v^2=g*1m v= sqrt(9.8) m/s
Now, find the momentum change
find the acceleration
vf^2=vi^2+2ad
d= distance for each stop.
vi=sqrt(9.8) m/s
vf=0
a= 1/2 (9.8)distance= 4.9 m/s^2/distance