What is the instantaneous velocity of an object moving in accordance to # f(t)= (e^(t-t^2),2t-te^t) # at # t=-1 #?

1 Answer
Yahia M.
Apr 8, 2018

#v=3.5791#(Length unit/unit of time)
and it's direction #theta=253.7756#degrees

Explanation:

#t# can't be negative as the time can't be negative but the following answer demonstrates the way to solve this kind of equations for #t=1# instead of #t=-1# and I will put the answer on double check.

This function is a displacement function for a particle moving in the Cartesian plane:
The function is divided into two parts #x,y# which are both parametric equations
so You can differentiate each part alone to get the velocity-time function
#f'(t)=((1-2t)e^(t-t^2),2-e^t-te^t)#

and by substituting with the value #t=-1# You get:

#f'(1)=((-1)e^0,2-e^1-(1)e^1)#
and by simplification You get:
#f'(-1)=(-1,2(1-e))#
so the magnitude of the velocity of the particle at that moment will be:
#v#=#sqrt((x^o)^2+(y^0)^2)#

and it's direction will be #tantheta=(y^0)/(x^0)#

And by the substituting, You get #v=3.5791#(Length unit/unit of time)
and #theta=253.775degrees#

I hope this was helpful.

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