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Q66P

Expert-verifiedFound in: Page 36

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**In a forward punch in karate, the fist begins at rest at the waist and is brought rapidly forward until the arm is fully extended. The speed ${\mathbf{v}}{\mathbf{\left(}}{\mathbf{t}}{\mathbf{\right)}}$****of the fist is given in figure for someone skilled in karate. The vertical scaling is set by ${{\mathbf{v}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{8}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\hspace{0.33em}}}{\mathbf{m}}{\mathbf{/}}{\mathbf{s}}{\mathbf{.}}$ **. **How far has the first moved at (a) time ${\mathbf{t}}{\mathbf{=}}{\mathbf{50}}{\mathbf{}}{\mathbf{ms}}$ **** (b) when the speed of the fist is maximum?**

(a) Position of a fist at a certain time is 0.13 meters

(b) Distance when the speed of fist is maximum 0.50 meters** **

The graph of v (m/s) vs t (s) with the vertical scale set by ${\mathrm{v}}_{\mathrm{s}}=8.0\mathrm{m}/\mathrm{s}.$ .

**The calculation of the region bounded by the curve within the given graph is called the area under the curve. The area under the curve will give the distance covered by the fist.**

We need to divide that area into two parts, to compute the position of the object at ${\mathrm{t}}_{1}=50\mathrm{ms}$,

The first part is from.0 to 10 ms The area has the shape of a triangle so the area under the curve is,

$\mathrm{Area}\mathrm{of}\mathrm{triangle}\left(\mathrm{A}\right)=0.01\mathrm{meters}$

2^{nd} part is from 10 to 50 ms that area has the shape of a trapezoid, so the area under the curve is,

$\mathrm{Area}\mathrm{of}\mathrm{trapezoid}\left(\mathrm{B}\right)=\frac{1}{2}\times 0.040\mathrm{sec}\times (2+4)\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}=0.12\mathrm{m}$

Now, $\mathrm{Area}\mathrm{of}\mathrm{triangle}+\mathrm{Area}\mathrm{of}\mathrm{trapezoid}=0.13\mathrm{m}$

Hence, the fist is at 0.13 m distance at 0.050 sec

From the graph, we conclude the following statement

The maximum speed is at $\mathrm{t}=0.120\mathrm{sec}.$

We need to find the area under the curve for regions between 0.050 sec to 0.090 sec and 0.090 sec to 0.120 sec.

Now, the area under the curve between 0.050 to 0.090 sec. is shaped as a trapezoid is,

$\mathrm{Area}\mathrm{of}\mathrm{trapezoid}\left(C\right)=\frac{1}{2}\times (0.040\mathrm{sec})\times (4+5\mathrm{m}/\mathrm{s})\phantom{\rule{0ex}{0ex}}=0.18\mathrm{m}$

The area under the curve between 0.090 to 0.120 sec is shaped as a trapezoid,

role="math" localid="1656158740938" $\mathrm{Area}\mathrm{of}\mathrm{trapezoid}\left(\mathrm{D}\right)=\frac{1}{2}\times (0.030\mathrm{sec})\times (5+7.5\mathrm{m}/\mathrm{s})\phantom{\rule{0ex}{0ex}}=0.19\mathrm{m}$

Therefore, the total area under the curve is $\left({A}_{T}\right)$,

${\mathrm{A}}_{\mathrm{T}}=\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{D}\phantom{\rule{0ex}{0ex}}=0.01+0.12+0.18+0.19\phantom{\rule{0ex}{0ex}}=0.50\mathrm{m}$

So, the distance covered by the fist at maximum speed is 0.50 m .

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