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?

Short answer

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

(b) The speed of torso is 3.90 m/s

Step-by-step solution

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.

Area under the curve

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

(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 $avst$

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.

(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$avst$.

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.