In this exercise, consider a particle moving on a circular path of radius b described by r(t) = b cos(ωt)i + b sin(ωt)j, where ω = du/dt is the constant angular velocity. find the acceleration vector and show that its direction is always toward the center of the circle.

Acceleration of the particle =

Step-by-step explanation:

We are given the position vector of a particle moving in a circle of radius b units.

r(t) = b cos(ωt)i + b sin(ωt)j

Velocity , v == -bω sin(ωt)i + bω cos(ωt)j

The magnitude of velocity, v =

Squaring both sides,

Since = 1

The acceleration towards the centre is called the centripetal acceleration and is given by

a =

a =

a =

answer: (-8,4) are the coordinates of a

step-by-step explanation:

let the side of square abcd is x .

thus the area of square abcd is .

now, by the diagram area of triangle ead= 1/2 × ea × ad = 1/2 × ea × x

( because ad is the side of square)

and, according to the question,

the areas of square abcd and right ead are equal.

⇒

⇒ea=2x

thus, ea: ab=2: 1

therefore, point a divides line segment eb into the ratio 2: 1.

⇒ by sectional formula coordinates of a are,

=

=

=(-8, 4)

thus the coordinates of a are (-8,4)

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step-by-step explanation: