Question

When a soccer ball is kicked toward a player and the player deflects the ball by “heading” it, the acceleration of the head during the collision can be significant. Figure 2-38 gives the measured acceleration a(t) of soccer player’s head for a bare head and a helmeted head, starting from rest. The scaling on the vertical axis is set by${\mathbf{a}}_{\mathbf{s}}\mathbf{=}\mathbf{200}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{.}\mathbf{At}\mathbf{}\mathbf{time}\mathbf{}\mathbf{t}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{\hspace{0.33em}}\mathbf{ms}$ , what is the difference in the speed acquired by the bare head and the speed acquired by the helmeted head?

Short answer

The difference in the speed acquired by the bare and head the speed acquired by the helmeted head is 0.56 m/s .

Step-by-step solution

Given data

The graph of $\mathrm{a}(\mathrm{m}/{\mathrm{s}}^2)$ against $\mathrm{t}(\mathrm{ms})$ for bare head and helmet is plotted. The scaling on the vertical axis is set by ${\mathrm{a}}_{\mathrm{s}}=200\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2$.

Area under the curve

The calculation of the region bounded by the curve within the given graph is called the area under the curve.The velocity can be determined by the graphical integration of acceleration versus time.

Calculation for the area under the cure for VBare

Divide the curve into 4 different parts:

From 0 to 2 ms, has the shape of a triangle, with the area

$$\begin{gathered} {\mathrm{Area}}_{0-2}=\frac{1}{2}\times (2\times {10}^{-3})\times 120 \\ =0.12\mathrm{m}/\mathrm{s} \end{gathered}$$

From 2 ms to 4 ms , has the shape of trapezoidal, with the area

$$\begin{gathered} {\mathrm{Area}}_{2-4}=\frac{1}{2}\times \mathrm{base}\times (\mathrm{sum}\mathrm{of}\mathrm{parallel}\mathrm{sides}) \\ =\frac{1}{2}\times (2\times {10}^{-3})\times (120+140) \\ =\frac{1}{2}\times (2\times {10}^{-3})\times 260 \\ =0.26\mathrm{m}/\mathrm{s} \end{gathered}$$

From 4 to 6 ms , has the shape of trapezoidal, with area

localid="1656305016065" $$\begin{gathered} {\mathrm{Area}}_{4-6}=\frac{1}{2}\times \mathrm{base}\times (\mathrm{sum}\mathrm{of}\mathrm{parallel}\mathrm{sides}) \\ =\frac{1}{2}\times (2\times {10}^{-3})\times (140+200) \\ =\frac{1}{2}\times (2\times {10}^{-3})\times 340 \\ =0.34\mathrm{m}/\mathrm{s} \end{gathered}$$

From 6 to 7 ms, has the shape of triangle, with area

$$\begin{gathered} {\mathrm{Area}}_{6-7}=\frac{1}{2}\times B\mathrm{ase}\times Height \\ =\frac{1}{2}\times (1\times {10}^{-3})\times 200 \\ =0.10\mathrm{m}/\mathrm{s} \end{gathered}$$

 Step 4: Calculation for area under the cure for VHelmeted

From 0 to 3 ms , has the shape of triangle, with area

$$\begin{gathered} {\mathrm{Area}}_{0-3}=\frac{1}{2}\times \mathrm{base}\times \mathrm{Height} \\ =\frac{1}{2}\times (3\times {10}^{-3})\times 40 \\ =0.060\mathrm{m}/\mathrm{s} \end{gathered}$$

From 3 to 4 , has the shape of rectangle, with area

$$\begin{gathered} {\mathrm{Area}}_{3-4}=\mathrm{base}\times \mathrm{height} \\ =(1\times {10}^{-3})\times 40 \\ =0.040\mathrm{m}/\mathrm{s} \end{gathered}$$

From 4 to 6 , has the shape of the trapezoidal, with area

$$\begin{gathered} {\mathrm{Area}}_{4-6}=\frac{1}{2}\times \mathrm{base}\times \left(\mathrm{sum}\mathrm{of}\mathrm{parallel}\mathrm{sides}\right) \\ =\frac{1}{2}\times (2\times {10}^{-3})\times (40+80) \\ =\frac{1}{2}\times (2\times {10}^{-3})\times 120 \\ =0.12\mathrm{m}/\mathrm{s} \end{gathered}$$

From 6 to 7 ms , has the shape of triangle, with area

$$\begin{gathered} {\mathrm{Area}}_{6-7}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ =\frac{1}{2}\times (1\times {10}^{-3})\times 80 \\ =0.040\mathrm{m}/\mathrm{s} \end{gathered}$$

Total area under the bare curve will be,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{Bare}}={\mathrm{Area}}_{0-2}+{\mathrm{Area}}_{2-4}+{\mathrm{Area}}_{4-6}+{\mathrm{Area}}_{6-7} \\ =0.12+0.26+0.34+0.10 \\ =0.82\mathrm{m}/\mathrm{s} \end{gathered}$$

Total area under the helmeted curve will be,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{helm}\mathrm{eted}}={\mathrm{Area}}_{0-3}+{\mathrm{Area}}_{3-4}+{\mathrm{Area}}_{4-6}+{\mathrm{Area}}_{6-7} \\ =0.060+0.040+0.12+0.040 \\ =0.26\mathrm{m}/\mathrm{s} \end{gathered}$$

The difference between the two curves will be,

$$\begin{gathered} ∆\mathrm{V}={\mathrm{V}}_{\mathrm{Bare}}-{\mathrm{V}}_{\mathrm{helmeted}} \\ =0.82-0.26 \\ =0.56\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the difference between the speeds will be 0.56 m/s .