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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

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 tor at least 1 ms will cause injury. Suppose; a small child rolls off a bed that is 0.40 m above the floor. If the floor is hardwood, the child’s head is brought to rest in approximately 2.0 min. 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.

To determine
Magnitude and duration of deceleration in both cases, to determine the risk of injury.

Explanation

Section 1:

To determine: The deceleration of the child.  

Answer: The deceleration of the child’s head after the impact is 2.0×103m/s2 .

Explanation:

Given Info:

The initial velocity of the child is 0 .

The acceleration of the child before the impact is 9.80m/s2 .

The falling distance is 0.40m .

The displacement of the child’s head after impact is 2.0×103m .

The final velocity of the child’s head after the impact is 0 .

The formula used to calculate the velocity of the child’s head just before impact is,

v=v02+2aΔy

  • Δy is the displacement of the child’s head
  • v is the final velocity of the child’s head
  • v0 is the initial velocity of the child’s head
  • a is the acceleration of the child’s head

Substitute 0 for v0 , 9.80m/s2 for a and 0.40m for Δy to find v .

v=(0)2+2(9.80m/s2)(0.40m)=2.8m/s

Thus, the final velocity of the child’s head just before impact is 2.8m/s .

The formula used to calculate the acceleration of the child’s head when it has an extra displacement after the impact is,

a'=vf2v22Δyi

  • Δyi is the displacement of the child’s head after impact
  • vf is the final velocity of the child’s head after impact
  • v is the initial velocity of the child’s head after impact
  • a' is the deceleration of the child’s head after impact

Substitute 0 for vf , 2.8m/s for v and 2.0×103m for Δyi to find a' .

a'=(0)2(2.8m/s)22(2.0×103m)=2.0×103m/s2

Thus, the deceleration of the child’s head after the impact is 2.0×103m/s2 .

Conclusion:

The deceleration of the child’s head after the impact is 2.0×103m/s2 .

Section 2:

To determine: The time taken for the child to decelerate.

Answer: The time taken for the child to decelerate is 1.4ms .

Explanation:

The formula used to calculate the time taken for the child to decelerate is,

t=2Δyiv

  • t is the time taken for the child to decelerate

Substitute 2.0×103m for Δyi and 2.8m/s for v to find t .

t=2(2.0×103m)2.8m/s=1.4×103s=(1

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