Traumatic brain injury such as concussion results when the head undergoes a very

Question

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 0.39 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 2.2 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.


hardwood floor magnitude_______ m/s2
hardwood floor duration________ ms
carpeted floor magnitude________ m/s2
carpeted floor duration________ms

Explanation / Answer

final velocity by using the Vf^2 =2as + Vi^2 equation so we plug in: Vf^2= 2(9.8m/s^2)(0.43m) + (0m/s)^2 and we get 2.9m/s next we calculate the deceleration of each material. since we know the initial and final velocity of the boy (initial is 2.9m/s, final is 0 because he comes to a stop) and we know the distance for each material, then we can solve this problem Vf^2 = 2as +Vi^2 for the hardwood floor (0 m/s)^2 = 2(a)(0.0019 m) +(2.9m/s)^2 using significant figures, the hardwood floor has a deceleration rate of approximately -2200m/s^2 to calculate the time, we can use the average velocity formula, S = ((vf+vi)/2)t plugging in (0.0019m) = ((0 +2.9m/s)/2)t we get the time to be 0.0013 s or 1.3ms for the carpeted floor we must do the same procedure. Vf^2 = 2as +Vi^2 ( 0 m/s)^2 = 2a(0.012m) + (2.9 m/s)^2 we get our deceleration to be -350 m/s^2 now for the time, (0.012m) = (0 + 2.9)/2 *t we get the time to be 8.2 ms

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