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/wikipediaupload.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 "