What is the difference between t = 0 and t = 1 seconds?
2 Answers
Given displacement vector
x=4cos(pit+pi/4)
Value of
x(0)=4cos(pixx0+pi/4)
=>x(0)=4cos(pi/4)
Value of
x(1)=4cos(pixx1+pi/4)
=>x(1)=4cos((5pi)/4)
Displacement between
Deltax=x(1)-x(0)
=>Deltax=4cos((5pi)/4)-4cos((5pi)/4)
=>Deltax=4[cos((5pi)/4)-cos((pi)/4)]
Making use of the unit circle reproduced below to rewrite in vector form. The first value in the round brackets represents
upload.wikimedia.org/wikipedia
=>Deltax=4[-sqrt2/2hatj-sqrt2/2hatj]
=>Deltax=-4sqrt2hatj
This is 1-D motion, along the number line:
-
x(t) = 4 cos (pi t + pi / 4) -
{(x(0) = 4 cos (pi /4) = 4/sqrt2),(x(1) = 4 cos ((5pi) /4) = - 4/sqrt2 ):}