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 1,000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.47 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 1.3 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

the final velocity, or the velocity of the child as he reaches the floor. since we know the childs acceleration due to gravity, we can determine his final velocity by using the Vf^2 =2as + Vi^2 equation

so we plug in: Vf^2= 2(9.8m/s^2)(0.47m) + (0m/s)^2

and we get 3.03 m/s

next we calculate the deceleration of each material.

since we know the initial and final velocity of the boy (initial is 3.03 m/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)(0.0021 m) +(3.03m/s)^2

using significant figures, the hardwood floor has a deceleration rate of -2185.92 m/s^2

to calculate the time, we can use the average velocity formula,

S = ((vf+vi)/2)t

plugging in (0.0021m) = ((0 +3.03m/s)/2)t

we get the time to be 0.001386 s or 1.386 ms

for the carpeted floor we must do the same procedure.

Vf^2 = 2as +Vi^2

( 0 m/s)^2 = 2a(0.013m) + (3.03 m/s)^2

we get our deceleration to be -353.11 m/s^2

now for the time,

(0.013m) = (0 +3.03)/2 *t

we get the time to be 8.85 ms

so here are our answers

hardwood floor deceleration: -2185.92 m/s^2

hardwood floor duration: 1.386 ms

carpeted floor deceleration: -353.11 m/s^2

carpeted floor duration : 8.85 ms

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