Question

The position of a particle is given in cm by x = (7) cos 9?t, where t is in seconds.

(a) Find the maximum speed.
...... m/s

(b) Find the maximum acceleration of the particle.
...... m/s²

(c) What is the first time that the particle is at x = 0 and moving in the +x direction?
....... s

Answer 1

Compare it with X=ACos(Wt)

angular frequency

W=9pi rad/s

Amplitude

A=7cm

a)

Maximum speed

V_max=AW=0.07*9pi=1.98 m/s

b)

Maximum accleration

a_max=W²A=(9pi)²*0.07=55.96 m/s²

c)

When X=0

0=7Cos(9pit)

9pit=Cos^-1(0)=3pi/2

t=0.1667 s

Answer 2

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

The position of a particle is given in cm by x = (4) cos 6pt, where t is in seconds.
(a) Find the maximum speed.(b) Find the maximum acceleration of the particle.
(c) What is the first time that the particle is at x = 0 and moving in the +x direction?   

Answer

Given that,
Position of particle x = 4 cos 6?t

If we compare this equation with x = A cos ?t , we get

Amplitude A = 4 cm

                 A  = 0.04 m

Angular frequency ? = 6? rad/s

a) Maximum speed V_max = ?A

                       

                                       =( 6? rad/s)( 0.04 m)

                              V_max = 0.7536 m/s

b) Maximum acceleration     a_max   = ?² A

                 

                                                    = (  6? rad/s )²   (0.04 m)

                       

                                             a_max = 14.19 m/s²

c) When    x = 0

        

           4 cos 6?t = 0

           

             cos 6?t = 0

               

              cos 6?t = cos 90⁰    

             

                    6?t = 90

                       

                        t = ( 90 / 6? )

                        t = ( 90 / 18.84 )

                        t = 4.77 s

Answer 3

x = 7(cos 9?t)

<< Find the maximum speed. >>

Differentiating the given function,

dx/dt = velocity = - 49? (sin 9?t)

and the maximum speed is when

sin 9?t = - 1

9?t = arc sin -1 = 3?/2

and solving for "t",

t = 1/6

Substitute t = 1/6 in the above equation for dx/dt and this will give you the maximum speed. I trust that you can proceed with the actual calculations on your own.


<< Find the maximum acceleration of the particle. >>

The second derivative of the function,

d^x/dt^2 = acceleration = - 49? (sin 9?t) = - 343(?)^2 (cos 9?t)

and the maximum acceleration is when cos 9?t = -1.

To determine the maximum acceleration, follow the same steps above in determining the maximum velocity. Again, I trust that you can do this on your own as you can simply follow the above procedure.


<< What is the first time that the particle is at x = 0 and moving in the +x direction? >>

When x = 0, the given function becomes

0 = (7 cm) cos 9?t

and the above becomes

cos 9?t = 0

9?t = arc cos 0

9?t = ?/2

and solving for "t"

t = 1/18

Answer 4