Physics, 14.09.2021 01:10 zahradawkins2007

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 0.50 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.9 mm. If the floor is carpeted, this stopping distance is increased to about 1.2 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.

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Answer from: mmczora22

Acceleration:

The rate of change of velocity with respect to time is known as acceleration. The SI unit of acceleration is $m/8^{2}$

Answer and Explanation:

Given:

. Distance of the bed from the floor: 0.48 m

. Stopping distance by hardwood: 1.7 mm

. Stopping distance by carpet: 1.1 cm

First, we will find the velocity of the child when they touch the floor using the formula:

. $v^{2}$ = $u^{2}$ +2gh

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

As the child is falling from the bed (from rest), the initial velocity will be 0.

So, we can write:

$v^{2} = (0)^{2} + 2gh$

$v^{2} = 2gh$

$v=\sqrt{2gh}$

$v=\sqrt{2gh} \\=\sqrt{2x9.8x0.48} \\=3.067 m/s$

Now, we will find the deceleration provided by the hardwood in stopping the child using the formula:

$vx^{2} =u^{2} -2aS$

where a is the deceleration and S is the distance traveled before stopping.

Since the child will come at a stop the final velocity will be 0, the initial velocity will be the velocity gained by the child while falling that we found in the first step, and the distance S for a stop is provided in the question.

$v^{2}= u^{2} -2aS\\0^{2} =(3.067^{2} )-2a(1.7x10x^{-3} )\\$

From the calculated deceleration we will find the time taken for the child to stop after hitting the hardwood using the formula:

v=u-at

where t is the time taken.

here the final velocity will be 0, as the child comes to rest, the initial velocity will be the same as calculated in the first part and the deceleration will be the value that we calculated in the second part.

$v=u-at\\v-u\\ --- = t\\-a$

$0-3.067\\ ---\\ -2766.61$ =t

$t=1.10x10^{-3}$

t=1.1ms

Now, we will use the same steps that we did to find the deceleration provided by the hardwood, but this time for the distance S we will use the value for the carpet provided in the question.

$v^{2} =u^{2} -2aS$

$0^{2} =(3.067^{2} ) -2a(1.1x10 -2)$

$a= 427.57 m/s^{2}$

Time for stopping by the carpet:

$v=u-at\\v-u\\---\\ -a$ = t

$0-3.067\\----\\-427.57$ =t

$t=7.17x10^{-3}$

t=7.17ms

According to the question when acceleration is greater than $1000 m/s^{2}$ for 1ms then it can cause injury. Since we calculated and observed that the acceleration for hardwood is greater than $1000 m/s^{2}$ and also lasted longer than 1 ms, we can say that the risk of injury on hardwood is greater than that of carpet. While the carpet also had a stopping time of greater than 1 ms, the acceleration was far less than $1000 m/s^{2}$ .

Hope this helps!

Answer from: Quest

grav forces affect all objects magnetic forces affect magnetic objects and electric currents

Answer from: Quest

As newton's first law states a object in motion stays in motion

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