The tendon from Lissa’s knee extensor muscles attaches to the tibia bone 1.5 in. (4 cm) below the center of her knee joint, and her foot is 15 in. (38 cm) away from her knee joint. What arc length does Lissa’s foot move through when her knee extensor muscles contract and their point of insertion on the tibia moves through an arc length of 2 in. (5 cm)?

s₂ = 0.475 m = 47.5 cm

Explanation:

The arc length and the angle of rotation are related through the formula:

s = rθ

where,

s = arc length

r = radius of curvature

θ = angle of rotation

First, we consider the arc length covered by the point of insertion of extensor muscles.

s₁ = r₁θ

where,

s₁ = arc length covered by insertion of extensor muscle = 5 cm

r₁ = length of insertion from knee = 4 cm

θ = Angle of Rotation = ?

Therefore,

5 cm = (4 cm)(θ)

θ = (5 cm)/(4 cm)

θ =  1.25 rad

Now, we consider the arc length covered by the foot.

s₂ = r₂θ

where,

s₂ = arc length covered by the foot = ?

r₂ = distance from knee to foot = 38 cm = 0.38 m

The angle of rotation will be the same for the foot as the insertion.

Therefore,

s₂ = (0.38 m)(1.25 rad)

s₂ = 0.475 m = 47.5 cm

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.

prndulum needs thread, weight clock to time swings