The equation r(t)=cos(5t)i + sin(5t)j, 0t≥0 describes the motion of a particle moving along the unit circle. answer the following questions about the behavior of the particle. a. does the particle have constant speed? if so, what is its constant speed? b. is the particle's acceleration vector always orthogonal to its velocity vector? c. does the particle move clockwise or counterclockwise around the circle? d. does the particle begin at the point left parenthesis 1 comma 0 right parenthesis(1,0)?

a) 5 units/s

b) yes

c) counter-clockwise

d) yes

Step-by-step explanation:

part a

r(t) = cos (5t) i + sin (5t)j

v(t) = dr(t) / dt = -5sin(5t) i + 5cos(5t)j

Hence, the particle has a constant speed of 5 units/s

part b

a(t) = dv(t) / dt = -25cos(5t) i - 25sin(5t)j

To check orthogonality of two vectors their dot product must be zero

a(t) . v(t) = (-25cos(5t) i - 25sin(5t)j) . (-5sin(5t) i + 5cos(5t)j)

= 125cos(5t)*sin(5t) -125cos(5t)*sin(5t)

= 0

Yes, the particles velocity vector is always orthogonal to acceleration vector.

part c

Use any two values of t and compute results of r(t)

t = 0 , r(0) = 1 i

t = pi/2, r(0) = j

Hence we can see that the particle moves counter-clockwise

part d

Find the value r(t) at t=0

r(0) = cos (0) i + sin (0) j

r(0) = 1 i + 0 j

Yes, the particle starts at point ( 1, 0)

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