Chapter 2, Problem 49P

### College Physics

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

Chapter
Section

### 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. Â

Answer: The deceleration of the childâ€™s head after the impact is 2.0Ã—103â€‰m/s2 .

Explanation:

Given Info:

The initial velocity of the child is 0 .

The acceleration of the child before the impact is âˆ’9.80â€‰m/s2 .

The falling distance is âˆ’0.40â€‰m .

The displacement of the childâ€™s head after impact is 2.0Ã—10âˆ’3â€‰m .

The final velocity of the childâ€™s head after the impact is 0 .

The formula used to calculate the velocity of the childâ€™s head just before impact is,

v=v02+2aÎ”y

• Î”y is the displacement of the childâ€™s head
• v is the final velocity of the childâ€™s head
• v0 is the initial velocity of the childâ€™s head
• a is the acceleration of the childâ€™s head

Substitute 0 for v0 , âˆ’9.80â€‰m/s2 for a and âˆ’0.40â€‰m for Î”y to find v .

v=(0)2+2(âˆ’9.80â€‰m/s2)(0.40â€‰m)=âˆ’2.8â€‰m/s

Thus, the final velocity of the childâ€™s head just before impact is âˆ’2.8â€‰m/s .

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

a'=vf2âˆ’v2âˆ’2Î”yi

• Î”yi is the displacement of the childâ€™s head after impact
• vf is the final velocity of the childâ€™s head after impact
• v is the initial velocity of the childâ€™s head after impact
• a' is the deceleration of the childâ€™s head after impact

Substitute 0 for vf , âˆ’2.8â€‰m/s for v and 2.0Ã—10âˆ’3â€‰m for Î”yi to find a' .

a'=(0)2âˆ’(âˆ’2.8â€‰m/s)2âˆ’2(2.0Ã—10âˆ’3â€‰m)=2.0Ã—103â€‰m/s2

Thus, the deceleration of the childâ€™s head after the impact is 2.0Ã—103â€‰m/s2 .

Conclusion:

The deceleration of the childâ€™s head after the impact is 2.0Ã—103â€‰m/s2 .

Section 2:

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

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

Explanation:

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

t=âˆ’2Î”yiv

• t is the time taken for the child to decelerate

Substitute 2.0Ã—10âˆ’3â€‰m for Î”yi and âˆ’2.8â€‰m/s for v to find t .

t=âˆ’2(2.0Ã—10âˆ’3â€‰m)âˆ’2.8â€‰m/s=1.4Ã—10âˆ’3â€‰s=(1

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started