THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS ARE PERFORMED, WHAT IS THE TOTAL ANGULAR DISTANCE AND THE TOTAL ANGULAR DISPLACEMENT UNDERGONE AT THE KNEE?

angular distance = 1700° and 29.7 rad

      also the angular displacement = 0

Explanation:

To explain this, i will give a breakdown of this works.

we are asked to find both the angular distance and displacement the knee undergo.

Ok to get the distance of the knee, we would first take note that for one to squat down and get back up, the knee would travel through 85° of flexion to gpo down, and also through another 85° of extension to return standing (upright). So, the actual angular distance of the squat is 170°.

taking ten squats, the knee would have to go through 170° motion times 10 i.e;

10 * 170° = 1700°

Therefore the angulaar distance is 1700°

now converting this distance to radians since we will be required to have our answer in both degree and rad.

Given that 2pi = 360°, it means that one degree will give 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1170° * 2π/360°  = 29.7 rad

∅ (rad) = 29.7 rad

b. For the other part, let us remember that angular displacement is equal to angular distance divided by time, so the angular displacement displacement of the knee will be zero, because the knee's position at the final third will be the same as the initial position.

cheers i hope this helps

π

i would think klcio3

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im doing peak rn so sry if its wrong