Question

The position of a particle is given in cm by x = (2) cos 9?t, where t is in seconds.

(a) Find the maximum speed.

0.565 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

x = 2*cos (9*pi*t)

We know that

V = dx/dt = -2*9*pi*sin (9*pi*t)

Vmax = |-2*9*pi| = 56.55 cm/sec = 0.5655 m/sec

Now

a = dV/dt = -18*9*pi^2*cos (9*pi*t)

a_max = |-18*9*pi^2| = 1598.8 cm/sec^2 = 15.988 m/sec^2

Part C

at t = 0 sec, particle is at x = 2*cos 0 deg = 2 cm (At extreme position)

at t = 0 sec, velocity of particle is

V = -18*pi*sin 0 = 0 m/sec

and it will start moving in left, since at t = 0^{+}, V
is negative.

Now particle will return at x = 0 and will be moving to right (+x direction) after 3T/4 time

T = time period = 2*pi/w = 2*pi/(9*pi)

T = 2/9

So first time Particle will be moving to +x direction will be = 3T/4 = 3*(2/9)/4 = 0.167 sec

Please Upvote.

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