Q67P - University Physics with Modern Physics

Question

When a car is hit from behind, its passengers undergo sudden forward acceleration, which can cause a severe neck injury known as whiplash. During normal acceleration, the neck muscles play a large role in accelerating the head so that the bones are not injured. But during a very sudden acceleration, the muscles do not react immediately because they are flexible; most of the accelerating force is provided by the neck bones. Experiments have shown that these bones will fracture if they absorb more than 8.0 J of energy. (a) If a car waiting at a stoplight is rear-ended in a collision that lasts for 10.0 m/ s, what is the greatest speed this car and its driver can reach without breaking neck bones if the driver’s head has a mass of 5.0 kg (which is about right for a 70-kg person)? Express your answer in m/s and in mi/h. (b) What is the acceleration of the passengers during the collision in part (a), and how large a force is acting to accelerate their heads? Express the acceleration in $\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$ and in g’s.

Step-by-Step Solution

Answer

(a) The greatest speed of the car is 1.8m/s or 4.0 mi/h.

(b) The acceleration of the passengers is $180 \mathrm{m}.{\mathrm{s}}^{-2}$ or 18.4 g.

Step 1: Given data

Energy = 8.0 J   

 

Collision = $10.0 \mathrm{m}.{\mathrm{s}}^2$ 

Step 2: Solution

Use the law of conservation energy to calculate the greatest speed of the car.

According to the law of conservation of energy,

 

${\mathrm{E}}_{\mathrm{i}}={\mathrm{E}}_{\mathrm{f}}$ 

 

Here, ${\mathrm{E}}_{\mathrm{i}}$ is the initial energy and ${\mathrm{E}}_{\mathrm{f}}$ is the final energy. 

Step 3: (a) Find the greatest speed this car

According to the law of conservation of energy,

 

$\frac{1}{2}{\mathrm{mv}}_{\max }^2=8.0 J$

 

Rearrange the above equation for speed and substitute 5.0 kg for m in above equation as follows:

$$\begin{gathered} {\mathrm{v}}_{\max }=\sqrt{\frac{2\times \left(8.0 \mathrm{J}\right)}{5.0 \mathrm{kg}}} \\ =1.8\mathrm{m}/\mathrm{s}\left(\frac{2.23694\mathrm{mi}/\mathrm{h}}{1 \mathrm{m}/\mathrm{s}}\right) \\ =4.0 \mathrm{mi}/\mathrm{h} \end{gathered}$$ 

 

 

Therefore, the greatest speed is 1.8 m/s or 4.0 mi/h.

Step 4: (b) Find the acceleration of the car

According to the kinematic equation, the acceleration of the car is,

$\mathrm{a}=\frac{{\mathrm{v}}_{\max }-\mathrm{u}}{\mathrm{t}}$ 

 

Substitute 0m/s for u , 4.0 mi/h  for $v_{max}$ , and 10 ms for t in the above equation as follows: 

$$\begin{gathered} \mathrm{a}=\frac{\left(4.0 \mathrm{mi}/\mathrm{h}\right)\left({\displaystyle \frac{1 \mathrm{m}/\mathrm{s}}{2.23694 \mathrm{mi}/\mathrm{h}}}\right)}{10 \mathrm{ms}\left({\displaystyle \frac{{10}^{-3} \mathrm{s}}{1 \mathrm{ms}}}\right)} \\ =180 \mathrm{m}.{\mathrm{s}}^{-2} \end{gathered}$$ 

 

Therefore, the acceleration of the car is $180 \mathrm{m}.{\mathrm{s}}^{-2}$.

 

The acceleration can be written as follows:

$$\begin{gathered} \mathrm{a}=180 \mathrm{m}.{\mathrm{s}}^{-2}\left(\frac{\mathrm{g}}{9.8 \mathrm{m}.{\mathrm{s}}^{-2}}\right) \\ =18.4 \mathrm{g} \end{gathered}$$ 

 

Therefore, the acceleration in terms of  is 18.4 g.