Question

The position of a particle in cm is given by x = (3) cos 9?t, where t is in seconds.

Answer 1

x = 3(cos 9?t)

<< Find the maximum speed. >>

Differentiating the given function,

dx/dt = velocity = - 27? (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

v=27*pi=84.82 cm/s

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

The second derivative of the function,

d^x/dt^2 = acceleration = - 27? (sin 9?t) = - 243(?)^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 = (3cm) 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