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Traumatic brain Injury such as concussion results when the undergoes very large acceleration Generally an than 800 m/s ^ 2 lasting for any length of time will not , whereas an acceleration greater than 1, 000 m/s ^ 2 for at least 1 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/s^2Hardwood floor duration msCarpeted floor magnitude m/s^2Carpeted floor duration ms

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ANSWER:

Hardwood floor magnitude: 2193.15 m/s^2

Hardwood floor duration: 1.38 ms

Carpeted floor magnitude: 354.27 m/s^2

Carpeted floor duration: 8.56 ms

STEP-BY-STEP EXPLANATION:

We have the following information:

Distance of the bed from the floor: 0.47 m

Stopping distance by hardwood: 2.1 mm

Stopping distance by carpet: 1.3 cm

We must find the speed of the child when he hits the ground using the following formula:

[tex]v^2=u^2+2gh[/tex]

Where v is the final velocity, u is the initial velocity, g is the gravitational acceleration and h is the distance traveled.

Replacing and solving for v

[tex]\begin{gathered} v^2=0^2+2\cdot9.8\cdot0.47 \\ v=\sqrt[]{9.212} \\ v=3.035\text{ m/s} \end{gathered}[/tex]

We can find the deceleration provided by the hardwood when stopping the child using the formula:

[tex]v^2=u^2-2aS[/tex]

Since the child will stop, the final speed will be 0, the initial speed will be the speed gained by the child while falling that we found in the first step, and the distance S to stop is given in the question.

Replacing and solving for a:

[tex]\begin{gathered} 2.1\text{ mm }\cdot\frac{1\text{ m}}{1000\text{ mm}}=2.1\cdot10^{-3}m \\ 0^2=3.035^2-2\cdot a\cdot(2.1\cdot10^{-3}) \\ 2\cdot a\cdot(2.1\cdot10^{-3})=9.211 \\ a=\frac{9.211}{2\cdot(2.1\cdot10^{-3})} \\ a=2193.15\text{ }\frac{m}{s^2} \end{gathered}[/tex]

From the calculated deceleration, we can find the time it takes for the child to stop after hitting the hardwood using the formula:

[tex]\begin{gathered} v=u-at \\ \text{replacing and solving for t:} \\ t=\frac{v-u}{-a} \\ t=\frac{0-3.035}{-2193.15} \\ t=0.00138\text{ s} \\ t=1.38\text{ ms} \end{gathered}[/tex]

We repeat the same procedure but now with the data of the carpet, therefore:

[tex]\begin{gathered} 1.3\text{ cm }\cdot\frac{1\text{ m}}{100\text{ cm}}=1.3\cdot10^{-2}m \\ 0^2=3.035^2-2\cdot a\cdot(1.3\cdot10^{-2}) \\ 2\cdot a\cdot(1.3\cdot10^{-2})=9.211 \\ a=\frac{9.211}{2\cdot(1.3\cdot10^{-2})} \\ a=354.27\text{ }\frac{m}{s^2} \end{gathered}[/tex]

Time for stopping by the carpet:

[tex]\begin{gathered} t=\frac{v-u}{-a} \\ t=\frac{0-3.035}{-354.27} \\ t=0.00856\text{ s} \\ t=8.56\text{ ms} \end{gathered}[/tex]