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The position of an object moving along a line is given by #p(t) = sin(2t- pi /3) +2 #. What is the speed of the object at #t = (2pi) /3 #?

Answer 1

Charlotte Aaron

#v((2pi)/3)=-2#

#v(t)=d/(d t) p(t)#

#v(t)=d/(d t)(sin(2t-pi/3)+2)#

#v(t)=2*cos(2t-pi/3)#

#"for "t=((2pi)/3) rarr v((2pi)/3)=2*cos(2*(2pi)/3-pi/3)#

#v((2pi)/3)=2*cos((4pi)/3-pi/3)#

#v((2pi)/3)=2*cos pi#

#cos pi=-1#

#v((2pi)/3)=-2*1#

#v((2pi)/3)=-2#

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Answer 2

Amelia Adams

To find the speed of the object at ( t = \frac{2\pi}{3} ), we need to differentiate the position function ( p(t) ) with respect to time ( t ), and then evaluate it at ( t = \frac{2\pi}{3} ).

The derivative of ( p(t) = \sin(2t - \frac{\pi}{3}) + 2 ) with respect to ( t ) is:

[ \frac{dp}{dt} = 2\cos(2t - \frac{\pi}{3}) ]

Evaluating ( \frac{dp}{dt} ) at ( t = \frac{2\pi}{3} ):

[ \frac{dp}{dt}\Bigg|_{t=\frac{2\pi}{3}} = 2\cos(2(\frac{2\pi}{3}) - \frac{\pi}{3}) ] [ = 2\cos(\frac{4\pi}{3} - \frac{\pi}{3}) ] [ = 2\cos(\frac{3\pi}{3}) ] [ = 2\cos(\pi) ] [ = -2 ]

So, the speed of the object at ( t = \frac{2\pi}{3} ) is ( \boxed{-2} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.