Q65P - Fundamentals Of Physics

Question

Figure 2-15a gives the acceleration of a volunteer’s head and torso during a rear-end collision. At maximum head acceleration, what is the speed of (a) the head and (b) the torso?

Step-by-Step Solution

Answer

(a)  The speed of the head is  225 m/s.

(b)  The speed of torso is  3.90 m/s

Step 1: Given data

The graph of acceleration vs time is given. The value of acceleration is in m/s² and the value of time is in ms.

Step 2: Area under the curve

The calculation of the region bounded by the curve within the given graph is called the area under the curve.

Step 3: (a) Calculations for speed of the head

From the graph, we can determine the following values,

$\mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{triangle}=\mathrm{Maximum} \mathrm{acceleration} \mathrm{of} \mathrm{head}"=90\frac{\mathrm{m}}{{\mathrm{s}}^2}$ 

The time t₁, when the head starts to accelerate is,

 $$\begin{gathered} {\mathrm{t}}_1=110 \mathrm{ms} \\ =0.110\mathrm{s} \end{gathered}$$

The time t₂, when the acceleration of the head is maximum.

 $$\begin{gathered} {\mathrm{t}}_2=160 \mathrm{ms} \\ =0.160 \mathrm{s} \\ \mathrm{The} \mathrm{base} \mathrm{of} \mathrm{the} \mathrm{triangle} =(0.160-0.110) \\ =0.05 \mathrm{s} \end{gathered}$$

 

The equation for the velocity is,

  $\mathrm{V}=\mathrm{at}$                                              (i)                                                                                                                        

The velocity  $\left(at\right)$ can be calculated from the area under the curve (region A) of $avs t$

The shape of the region A is triangular.

So, the speed of the head at maximum head acceleration is,

 $$\begin{gathered} \mathrm{v}=\left(\frac{1}{2}\right)\left(0.160-0.110\right)\left(90\right) \\ =2.25 \mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the velocity of the head is calculated as  225 m/s.

Step 4: (b) Calculations for speed of torso

From the graph, find the following quantities,

 $\mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{triangle}=\mathrm{Maximum} \mathrm{acceleration} \mathrm{of} \mathrm{torso}=50\mathrm{m}/{\mathrm{s}}^2$

The time t₃ when the torso starts accelerating is,

 $$\begin{gathered} {\mathrm{t}}_3=40 \mathrm{ms} \\ =0.04 \mathrm{s} \end{gathered}$$

The time t₄ when the acceleration of the torso is maximum,

$$\begin{gathered} {\mathrm{t}}_4=100 \mathrm{ms} \\ =0.100 \mathrm{s} \\ \mathrm{The} \mathrm{base} \mathrm{of} \mathrm{triangle}=0.1-0.04 \\ =0.6 \mathrm{s} \end{gathered}$$ 

 

As velocity is defined as,  $\mathrm{v}=\mathrm{at}$  

The velocity  (at)  can be calculated from the area under the curve of $a vs t$ .

So, the speed of torso at maximum acceleration in this region is,

$$\begin{gathered} \mathrm{v}=\frac{1}{2} (0.1-0.04)\left(50\right) \hspace{0.33em}\hspace{0.33em}\hspace{0.33em} \\ =1.50" \mathrm{m}/\mathrm{s} \end{gathered}$$ 

Between 100 ms to 120 ms, the shape of the region B is rectangle. For this region the given values of quantities are as follows:

 $$\begin{gathered} \mathrm{Maximum} \mathrm{acceleration} \mathrm{of} \mathrm{torso}"=50\mathrm{m}/{\mathrm{s}}^2 \\ {\mathrm{t}}_5=100 \mathrm{ms} \\ =0.1 \mathrm{s} \\ {\mathrm{t}}_6=120 \mathrm{ms} \\ =0.120 \mathrm{s} \end{gathered}$$

 

 So, the speed of torso at maximum acceleration in this region is,    

 $\mathrm{v}=1.0 \mathrm{m}/\mathrm{s}$

From 120 ms to 160 ms, the shape is trapezoid, so the speed of torso in this region is,

 $$\begin{gathered} \mathrm{v}=\frac{1}{2} (0.0400)\times (50.0+20.0) \hspace{0.33em}\hspace{0.33em}\hspace{0.33em} \\ =1.40\hspace{0.33em}"\mathrm{m}/\mathrm{s}" \end{gathered}$$

The speed of torso at maximum acceleration is,

 $\mathrm{v}=1.5+1+1.4=3.90 \mathrm{m}/\mathrm{s}$

Hence the speed of torso is  3.9 m/s.