Chapter 2, Problem 49P

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.

Answer to Problem 49P

The deceleration of the child’s head after the impact is $2.0\times {10}^3{\text{m/s}}^2$ and the time taken for the child to decelerate is $1.4\text{ms}$. The deceleration of the child’s head after the impact in the carpeted floor is $3.9\times {10}^2{\text{m/s}}^2$ and the time taken for the child to decelerate in the carpeted floor is $7.1\text{ms}$.

Explanation of Solution

Section 1:

To determine: The deceleration of the child.  

Answer: The deceleration of the child’s head after the impact is $2.0\times {10}^3{\text{m/s}}^2$.

Explanation:

Given Info:

The initial velocity of the child is $0$.

The acceleration of the child before the impact is $-9.80{\text{m/s}}^2$.

The falling distance is $-0.40\text{m}$.

The displacement of the child’s head after impact is $2.0\times {10}^{-3}\text{m}$.

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,

$v=\sqrt{v_0^2+2a\Delta y}$

• $\Delta y$ is the displacement of the child’s head
• $v$ is the final velocity of the child’s head
• $v_0$ is the initial velocity of the child’s head
• $a$ is the acceleration of the child’s head

Substitute $0$ for $v_0$, $-9.80{\text{m/s}}^2$ for $a$ and $-0.40\text{m}$ for $\Delta y$ to find $v$.

$\begin{array}{c}v=\sqrt{{\left(0\right)}^2+2\left(-9.80{\text{m/s}}^2\right)\left(0.40\text{m}\right)}\\ =-2.8\text{m/s}\end{array}$

Thus, the final velocity of the child’s head just before impact is $-2.8\text{m/s}$.

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

$a\text{'}=\frac{v_f^2-v^2}{-2\Delta y_i}$

• $\Delta y_i$ is the displacement of the child’s head after impact
• $v_f$ is the final velocity of the child’s head after impact
• $v$ is the initial velocity of the child’s head after impact
• $a\text{'}$ is the deceleration of the child’s head after impact

Substitute $0$ for $v_f$, $-2.8\text{m/s}$ for $v$ and $2.0\times {10}^{-3}\text{m}$ for $\Delta y_i$ to find $a\text{'}$.

$\begin{array}{c}a\text{'}=\frac{{\left(0\right)}^2-{\left(-2.8\text{m/s}\right)}^2}{-2\left(2.0\times {10}^{-3}\text{m}\right)}\\ =2.0\times {10}^3{\text{m/s}}^2\end{array}$

Thus, the deceleration of the child’s head after the impact is $2.0\times {10}^3{\text{m/s}}^2$.

Conclusion:

The deceleration of the child’s head after the impact is $2.0\times {10}^3{\text{m/s}}^2$.

Section 2:

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

Answer: The time taken for the child to decelerate is $1.4\text{ms}$.

Explanation:

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

$t=\frac{-2\Delta y_i}{v}$

• $t$ is the time taken for the child to decelerate

Substitute $2.0\times {10}^{-3}\text{m}$ for $\Delta y_i$ and $-2.8\text{m/s}$ for $v$ to find $t$.

$\begin{array}{c}t=\frac{-2\left(2.0\times {10}^{-3}\text{m}\right)}{-2.8\text{m/s}}\\ =1.4\times {10}^{-3}\text{s}\\ \text{=}\left(1.4\times {10}^{-3}\text{s}\right)\left(\frac{1\text{ms}}{{10}^{-3}\text{s}}\right)\\ =1.4\text{ms}\end{array}$

Thus, the time taken for the child to decelerate is $1.4\text{ms}$.

Conclusion:

The time taken for the child to decelerate is $1.4\text{ms}$.

Section 3:

To determine: The deceleration of the child in the carpeted floor.

Answer: The deceleration of the child’s head after the impact in the carpeted floor is $3.9\times {10}^2{\text{m/s}}^2$.

Explanation:

Given Info:

The initial velocity of the child is $-2.8\text{m/s}$.

The displacement of the child’s head after impact in the carpeted floor is $1.0\times {10}^{-3}\text{m}$.

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

$a_c^{\text{'}}=\frac{v_f^2-v^2}{-2\Delta y_{ic}}$

• $\Delta y_{ic}$ is the displacement of the child’s head after impact in the carpeted floor
• $a_c^{\text{'}}$ is the deceleration of the child’s head after impact in the carpeted floor

Substitute $0$ for $v_f$, $-2.8\text{m/s}$ for $v$ and $1.0\times {10}^{-3}\text{m}$ for $\Delta y_{ic}$ to find $a_c^{\text{'}}$.

$\begin{array}{c}a\text{'}=\frac{{\left(0\right)}^2-{\left(-2.8\text{m/s}\right)}^2}{-2\left(1.0\times {10}^{-3}\text{m}\right)}\\ =3.9\times {10}^2{\text{m/s}}^2\end{array}$

Thus, the deceleration of the child’s head after the impact in the carpeted floor is $3.9\times {10}^2{\text{m/s}}^2$.

Conclusion:

The deceleration of the child’s head after the impact in the carpeted floor is $3.9\times {10}^2{\text{m/s}}^2$.

Section 4:

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

Answer: The time taken for the child to decelerate in the carpeted floor is $7.1\text{ms}$.

Explanation:

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

$t_c=\frac{-2\Delta y_{ic}}{v}$

• $t_c$ is the time taken for the child to decelerate

Substitute $1.0\times {10}^{-3}\text{m}$ for $\Delta y_{ic}$ and $-2.8\text{m/s}$ for $v$ to find $t_c$.

$\begin{array}{c}t=\frac{-2\left(1.0\times {10}^{-3}\text{m}\right)}{-2.8\text{m/s}}\\ =7.1\times {10}^{-3}\text{s}\\ \text{=}\left(7.1\times {10}^{-3}\text{s}\right)\left(\frac{1\text{ms}}{{10}^{-3}\text{s}}\right)\\ =7.1\text{ms}\end{array}$

Thus, the time taken for the child to decelerate in the carpeted floor is $7.1\text{ms}$.

Conclusion:

The time taken for the child to decelerate in the carpeted floor is $7.1\text{ms}$.