Traumatic brain injury such as concussionn results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s^2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s^2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is .40 m above the floor. If the floor is hardwood, the child's head is brought to rest in approxeimately 2.0 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.
Calculate the velocity of the body when it hits the floor.
V = sqrt (2gH) = 2.8 m/s
The time required needed to stop the fall in a distance d is given by
(V/2)*T = d since V/2 is the average velocity during deceleration.
T = 2d/V
acceleration rate = V/T = V^2/(2d)
= 392 m/s^2 for a carpeted floor
Compare the result for a hardwood floor.
To calculate the magnitude and duration of deceleration in both cases, we can use the kinematic equation for uniformly accelerated motion:
\(v_f^2 = v_i^2 + 2a d\)
Where:
\(v_f\) is the final velocity of the child's head (which is 0 in this case, as the child's head comes to rest),
\(v_i\) is the initial velocity of the child's head (which we need to calculate),
\(a\) is the deceleration,
\(d\) is the stopping distance.
Let's calculate for each case:
1. Case with a hardwood floor:
Given:
Stopping distance, \(d_1 = 2.0 \, \text{mm} = 0.002 \, \text{m}\)
Initial velocity, \(v_i\) is unknown
Using the kinematic equation:
\(0^2 = v_i^2 + 2a \cdot 0.002\)
We need one more equation to solve for \(a\) and \(v_i\). We know that acceleration is the rate of change of velocity with time:
\(a = \frac{{\Delta v}}{{\Delta t}}\)
Both \(\Delta v\) and \(\Delta t\) are unknowns. However, we can find a relationship between stopping distance and time by considering constant deceleration:
\(d_1 = \frac{{v_f^2}}{{2a}}\)
Now we can solve for \(a\) in terms of \(d_1\):
\(a = \frac{{v_f^2}}{{2d_1}} = \frac{{0^2}}{{2 \cdot 0.002}} = 0 \, \text{m/s}^2\)
Since the floor is hardwood, the stopping distance does not decelerate the child's head in this scenario. Therefore, no deceleration or risk of injury occurs.
2. Case with a carpeted floor:
Given:
Stopping distance, \(d_2 = 1.0 \, \text{cm} = 0.01 \, \text{m}\)
Initial velocity, \(v_i\) is unknown
Using the kinematic equation:
\(0^2 = v_i^2 + 2a \cdot 0.01\)
Using the relationship between stopping distance and deceleration:
\(a = \frac{{v_f^2}}{{2d_2}} = \frac{{0^2}}{{2 \cdot 0.01}} = 0 \, \text{m/s}^2\)
Similar to the previous case, the stopping distance due to the carpeted floor does not cause any deceleration or risk of injury.
In both cases, the magnitude of deceleration is 0 m/s², and as a result, there is no risk of injury to the child.