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Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 las

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asked 2021-01-22

Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than \(800 m/s^{2}\) lasting for any lengthof 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 0.42 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 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 lessthan the speed you calculate.

Answers (1)

2021-01-23

Determine the velocity of the child just as it hits the floor:
\(v(i) = 0\)
\(v(f) = ??\)
\(d = 0.37 m\)
\(a = 9.8 \displaystyle\frac{{m}}{{s}^{{2}}}\)
\(\displaystyle{v}{\left({f}\right)}^{{2}}={v}{\left({i}\right)}^{{2}}+{2}{a}{d}\)
\(\displaystyle{v}{\left({f}\right)}^{{2}}={0}+{\left({2}\right)}{\left({9.8}\right)}{\left({0.37}\right)}\to\) initial velocity iszero
\(v(f) = 2.7 m/s\)
Now, determine the acceleration upon impact with the floor under both hardwood and carpet, knowing that the final velocity will be zero:
Hardwood:
\(\displaystyle{v}{\left({f}\right)}^{{2}}={v}{\left({i}\right)}^{{2}}+{2}{a}{d}\)
\(\displaystyle{0}={2.7}^{{2}}+{\left({2}\right)}{\left({a}\right)}{\left({0.0018}\right)}\to\) \(1.8mm =0.0018m\)
\(\displaystyle{a}=-\frac{{2.7}^{{2}}}{{0.0036}}\)
\(\displaystyle{a}=-{2025}\frac{{m}}{{s}^{{2}}}=-{2}\times{10}^{{3}}\frac{{m}}{{s}^{{2}}}\) (negative acceleration for slowing)
Carpet:
\(\displaystyle{v}{\left({f}\right)}^{{2}}={v}{\left({i}\right)}^{{2}}+{2}{a}{d}\)
\(\displaystyle{0}={2.7}^{{2}}+{\left({2}\right)}{\left({a}\right)}{\left({0.01}\right)}\to\)\( 1cm =0.01m\)
\(\displaystyle{a}=-\frac{{2.7}^{{2}}}{{0.02}}\)
\(\displaystyle{a}=-{364.5}\frac{{m}}{{s}^{{2}}}=-{3.6}\times{10}^{{1}}\frac{{m}}{{s}^{{2}}}\) (negative acceleration for slowing)

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