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An object's two dimensional velocity is given by #v(t) = ( e^t-2t , 2t-4e^2 )#. What is the object's rate and direction of acceleration at #t=a #?

Answer 1

Isabella Alvarez

The rate of acceleration is #=sqrt(e^(2a)-4e^(a)+8)ms^-2# in the direction of #arctan(2/(e^(a)-2))#

The derivative of the velocity is the acceleration.

#v(t)=(e^t-2t, 2t-4e^2)#

#A(t)=v'(t)=(e^t-2, 2)#

Consequently,

#A(a)=(e^(a)-2, 2)#

The acceleration rate is

#||a(4)||=sqrt((e^(a)-2)^2+2^2)#

#=sqrt(e^(2a)-4e^(a)+4+4)#

#=sqrt(e^(2a)-4e^(a)+8)ms^-2#

The path is

#theta=arctan(2/(e^(a)-2))#

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Answer 2

Sebastian Acosta

The object's acceleration at ( t = a ) is ( a(t) = (1 - 2, 2) ) or ( a(t) = (-1, 2) ), depending on the value of ( a ). The rate of acceleration is (\sqrt{(-1)^2 + 2^2} = \sqrt{5}). The direction of acceleration is given by the unit vector ( \frac{a(t)}{|a(t)|} = \left(\frac{-1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right) ).

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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.