The distance s of an object from the origin at time t 0 (in seconds) is given. T

Question

The distance s of an object from the origin at time t 0 (in seconds) is given. The motion is along a horizontal line with the positive direction to the right.

s(t) = 3 cos(t),    0 t 2

(a) Determine the intervals during which the object moves to the right and the intervals during which it moves to the left. (Enter your answer using interval notation. If an answer does not exist, enter DNE.)

(b) When does the object reverse direction? (If an answer does not exist, enter DNE. Enter your answers as a comma-separated list.)

t =

(c) When is the velocity of the object increasing and when is it decreasing? (Enter your answer using interval notation. If an answer does not exist, enter DNE.)

moving to the right :    The distance s of an object from the origin at time t 0 (in seconds) is given. The motion is along a horizontal line with the positive direction to the right. s(e) 3 cos(t), 0 SL S 2 (a) Determine the intervals during which the object moves to the right and the intervals during which it moves to the left. (Lnter your answer using interval notation. If an answer does not exist, enter DNE.) moving to the right moving to the left (b) When does the object reverse direction? (If an answer does not exist, enter DNE. Enter your answers as a comma-separated lst.) (c) When is the velocity of the object increasing and when is it decreasing? (Enter your answer using interval notation. If an answer does not exist, enter DNE.) increasing

Explanation / Answer

s = 3cos(pi*t)
0 <= t <= 2

a) Right and left :
v = -3pisin(pi*t) = 0
sin(pi*t) = 0
t = 0 , 1 and 2

So, region 1 : (0,1)
Test = 0.5
WE find v = negative
So, leftward

Region 2 : (1,2)
Test = 1.5
We find v = -3pi*sin(pi*1.5) to be positive
So, rightward as v > 0

Right : (1,2)
Left : (0,1)

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b)
Reverse direction :
It reversed direction at t = 1

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c)
v = -3pisin(pi*t)

Deriving :
a = -3pi^2 * cos(pi.t) = 0
t = 1/2 and 3/2

Region 1 : (0,1/2)
Test = 1/4
We find a < 0

Region 2: (1/2 , 3/2)
Test = 1, we find a > 0

Region 3 : (3/2 , 2)
We find a< 0

So, vel increasing when a > 0
inc : (1/2 ,3/2)
dec : (0,1/2) U (3/2,2)

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