Please help me solve these problems

Question

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 0.54 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.4 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/s2
hardwood floor duration	ms
carpeted floor magnitude	m/s2
carpeted floor duration	ms

Answer 1

To solve this problem, we'll first find the velocity of the child's head just before it hits the floor using the conservation of mechanical energy. Then, we'll use the stopping distances for the hardwood and carpeted floors to calculate the deceleration magnitudes and durations.

1. Find the initial velocity just before hitting the floor:

Using the conservation of mechanical energy, we have:

mgh = 1/2 mv^2

v = √(2gh)

where m is the mass of the child (which will cancel out), g is the acceleration due to gravity (9.81 m/s^2), and h is the height (0.54 m). Substituting the values:

v = √(2 * 9.81 m/s^2 * 0.54 m) ≈ 3.27 m/s

2. Calculate deceleration magnitudes and durations for hardwood and carpeted floors:

For the hardwood floor:

Stopping distance, d₁ = 1.8 mm = 0.0018 m

Using the equation v² = u² + 2as, where v is the final velocity (0 m/s), u is the initial velocity (3.27 m/s), a is the deceleration, and s is the stopping distance:

0 = 3.27^2 + 2a * 0.0018

a = -3.27^2 / (2 * 0.0018) ≈ -2986 m/s^2 (magnitude)

To find the duration, use the equation v = u + at:

0 = 3.27 + (-2986) * t

t = 3.27 / 2986 ≈ 0.0011 s = 1.1 ms

For the carpeted floor:

Stopping distance, d₂ = 1.4 cm = 0.014 m

Using the equation v² = u² + 2as:

0 = 3.27^2 + 2a * 0.014

a = -3.27^2 / (2 * 0.014) ≈ -383 m/s^2 (magnitude)

To find the duration, use the equation v = u + at:

0 = 3.27 + (-383) * t

t = 3.27 / 383 ≈ 0.0085 s = 8.5 ms

Results:

hardwood floor magnitude: 2986 m/s^2

hardwood floor duration: 1.1 ms

carpeted floor magnitude: 383 m/s^2

carpeted floor duration: 8.5 ms

While both cases have decelerations higher than the 800 m/s² threshold, the hardwood floor case is above the critical 1000 m/s² value and lasts for more than 1 ms, which puts the child at a higher risk of injury. The carpeted floor, with a lower deceleration, would be a safer scenario.