What is the difference between t = 0 and t = 1 seconds?

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2 Answers
Aug 12, 2018

Given displacement vector

#x=4cos(pit+pi/4)#

Value of #x# at #t=0#

#x(0)=4cos(pixx0+pi/4)#
#=>x(0)=4cos(pi/4)#

Value of #x# at #t=1#

#x(1)=4cos(pixx1+pi/4)#
#=>x(1)=4cos((5pi)/4)#

Displacement between #t=0and t=1# is

#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 #cos# or #hatj# and second value represents #sin# or #hati# component for a particular angle.
upload.wikimedia.org/wikipedia

#=>Deltax=4[-sqrt2/2hatj-sqrt2/2hatj]#
#=>Deltax=-4sqrt2hatj#

Aug 12, 2018

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 ):}#

#Delta x = - 4/sqrt2 - 4/sqrt2 = - 8/sqrt2 =- 4 sqrt2 " metres " #