Question

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?

Short answer

(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

Step-by-step solution

Given data

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

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 area under the curve will give the distance covered by the fist.

(a) Calculations to find out the position of an object at any time to find the area under the curve:

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,

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{trapezoid}\left(\mathrm{B}\right)=\frac{1}{2}\times 0.040\sec \times (2+4)\mathrm{m}/\mathrm{s} \\ =0.12\mathrm{m} \end{gathered}$$

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

(b) Calculations for distance at which fist has maximum velocity:

From the graph, we conclude the following statement

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

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,

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{trapezoid}\left(C\right)=\frac{1}{2}\times (0.040\sec )\times (4+5\mathrm{m}/\mathrm{s}) \\ =0.18\mathrm{m} \end{gathered}$$

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

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

Therefore, the total area under the curve is $\left(A_T\right)$,

$$\begin{gathered} {\mathrm{A}}_{\mathrm{T}}=\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{D} \\ =0.01+0.12+0.18+0.19 \\ =0.50\mathrm{m} \end{gathered}$$

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