In a forward punch in karate, the fist begins at rest at the waist and is brough

Question

In a forward punch in karate, the fist begins at rest at the waist and is brought rapidly forward until the arm is full extended. The speed v(t) of the fist is given in the figure on the right for someone skilled in karate. The vertical scaling is set by vs = 8.0 m/s. By getting the equations for the v(t), estimate how far the fist has moved at time f = 50 ms? By using the graphical interpretation on the graph in the figure, estimate how far the fist has moved at time t = 50 ms? By using the graphical interpretation on the graph in the figure, estimate how far the fist has moved, when is the speed of the fist is maximum?.

Explanation / Answer

(1)it is at a far of distance=speed*time

          =4*50*10^-3

        =200mm

(2)The distance will beV(t) as area of upto 50ms

then area of triangle upto 10ms and area from 10ms to 50ms

then upto 10ms area is

=(bh)/2

=10*10^-3*2/2

=10*10^-3m

to 0-10m/s
10s* 2m/s * (1/2)= 10m

then area from 10 to 50 ms is

=(2*40*10^-3)/2+(40*10^-3*2)

=120*10^-3m

then it moves at a distance of

=120+10mm

=130mm

(3) It is maximum at 120 ms hence the total area upto 120ms is

area upto from 50ms to 90 ms is

area =4*40*10^-3+(1*40*10^-3)/2

=180*10^-3 m

area from 90ms to 120ms is

area =30*6*10^-3+30*2.5*10^-3/2

=217.5ms

total net far is

=217.5+180+130mm

=527.5 mm

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