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

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Answer 1 · 2018-05-07T14:03:12

#(-3pi)/2+2#

The position function of the object is given by:

#p(t)=2t-3cos(pi/2t)+2#

Since the velocity is displacement over time, it means that it the rate of changing the position over time, or the derivative of the function.

Then we got,

#v(t)=p'(t)#

Let's find the hardest part, which I think is:

#d/dt(3cos(pi/2t))#

We get:

#=3d/dt(cos(pi/2t))#

#=3d/dt(cos((pit)/2))#

Let #u=(pit)/2,:.(du)/dt=pi/2#.

Then, #y=cosu,:.dy/(du)=-sinu#.

Combine to get:

#=-sinu*pi/2#

#=-sin((pit)/2)*pi/2#

#=-pi/2sin((pit)/2)#

So,
#d/dt(3cos(pi/2t))=(-3pi)/2sin((pit)/2)#

And now, we get:

#p'(t)=2-[(-3pi)/2sin((pit)/2)]+0#

#=(3pi)/2sin((pit)/2)+2#

So, at #t=3#, we get:

#=(3pi)/2sin((3pi)/2)+2#

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