The velocity function of a moving particle on a coordinate line is v(t) = 3 cos(2t)    for 0 is less than or equal to t is greater than or equal to 2pi (a) Determine when the particle is moving...

The velocity function of a moving particle on a coordinate line is v(t) = 3 cos(2t)

 

 

for 0 is less than or equal to t is greater than or equal to 2pi

(a) Determine when the particle is moving to the right.

 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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Since the problem provides the equation of velocity, you need to solve for t the equation `v(t) = 0`  and then you need to discuss the sign of function, such that:

`3cos 2t = 0 => {(3!=0),(cos 2t = 0):} => 2t = +-pi/2 + 2n*pi `

`t = +-pi/4 + n*pi `

If `n = 0 => t = pi/4, t = pi-pi/4, t = 2pi-pi/4 `

`t = {pi/4 ;(3pi)/4 ; (7pi)/4}`

If `n = 1 => t = pi+pi/4 = 5pi/4 `

Since `t = 2pi+pi/4>2pi` , hence, the set of solutions to the equation `v(t) = 0`  is `t ={pi/4 ; (3pi)/4 ; (5pi)/4 ;(7pi)/4}.`

You need to evaluate the sign of function such that:

`t = pi/2 => 3cos((2pi)/2) = 3 cos pi = -3 < 0`

`t = pi => 3 cos 2pi = 3 > 0`

Hence, if `t in (0,pi/4)`  then the function `v(t)>0` , hence, the particle is moving to the right.

Hence, if `t in (pi/4,(3pi)/4)`  then the function `v(t)<0` , hence, the particle is moving to the left.

Hence, if `t in ((3pi)/4,(5pi)/4)`  then the function `v(t)>0` , hence, the particle is moving to the right.

Hence, if `t in ((5pi)/4,(7pi)/4)`  then the function `v(t)<0` , hence, the particle is moving to the left.

Hence, if `t in ((7pi)/4,2pi)`  then the function `v(t)>0,`  hence, the particle is moving to the right.

Hence, evaluating the intervals of values of`t` when the particle moves to the right, yields `t in (0,pi/4)U((3pi)/4,(5pi)/4)U((7pi)/4,2pi).`

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