The motion of an object is described by the equation below.?

Question

The motion of an object is described by the equation below.
x = (0.20 m) cos(t/6)

(a) Find the position of the object at t = 0 and at t = 0.40 s. t = 0 m
t = 0.40 s 0.99
Your answer differs from the correct answer by 10% to 100%. m
(b) Find the amplitude of the motion.
m
(c) Find the frequency of the motion.
Hz
(d) Find the period of the motion.
s

Answer 1

It has been a while since I did this kind of homework...
Okay
If remember correctly
X=Acos(wt+phase shift)
A- amplitude
w – angular frequency
also w=2 pi f
pi=3.14....
f – the frequency
x=(0.20 m) cos(t/6)

(a)at t=0 x(0) =(0.20) cos(0/6)
x(0)=.20 m
at t=0.40sec
x(.4)=.20cos(.4/6)=.1999m and so on

(b) A=.2m
(c) w=2 pi f =1/6
Then f=2 pi/6=1.047Hz
(d) T=1/f
So T=1/1.047Hz=0.955 sec

Answer 2

If the associated fee is v then v^2 =(dx/dt)^2+(dy/dt)^2 =(2t-3)^2+(3t^2-2)^2 =0 while the two 2t-3=0 and 3t^2-2=0 yet there is not any such value of t. there'll be no vertical action while dy/dt=0 giving 3t^2-2=0 giving t=?(2/3), assuming t>0, giving x=2/3 - 3?(2/3) and y= (2/3)?(2/3)-2?(2/3)=-(4/3)?(2/3). there'll be no horizontal action while dx/dt=0 giving 2t-3=0, t=3/2 and x=-9/4, y=3/8. you could desire to try this employing vectors. the placement vector of the article is given by employing r=(t^2-3t)i+(t^3-2t)j the associated fee vector is dr/dt =(2t-3)i+(3t^2-2)j dr/dt=0 while 2t-3=0 and 3t^2-2=0 as previously.