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An object's two dimensional velocity is given by #v(t) = (t^2 -t +1 , t^3- 3t )#. What is the object's rate and direction of acceleration at #t=3 #?

Answer 1

Elena Albarran

The rate of acceleration is #=24.52ms^-2# in the dorection #=78.2^@# anticlockwise from the x-axis

The derivative of velocity is the acceleration.

#v(t)=(t^2-t+1, t^3-3t)#

#a(t)=v'(t)=(2t-1,3t^2-3)#

At #t=3#, the acceleration is

#a(3)=(5,24)#

The acceleration rate is

#=||a(3)||=sqrt(5^2+24^2)=sqrt601=24.52ms^-2#

The path is

#theta=arctan(24/5)=78.2^@#

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Answer 2

Elena Albarran

The object's acceleration at t=3 is (4, 18). The rate of acceleration is 2 units per second squared, and the direction is along the x-axis and y-axis.

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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.