An object's two dimensional velocity is given by #v(t) = ( 5t-t^3 , 3t^2-2t)#. What is the object's rate and direction of acceleration at #t=4 #?
Direction:
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To find the acceleration, take the derivative of the velocity function with respect to time:
a(t) = v'(t) = (5 - 3t^2, 6t - 2)
Then, plug in t = 4:
a(4) = (5 - 3(4)^2, 6(4) - 2) = (5 - 48, 24 - 2) = (-43, 22)
The magnitude of acceleration is given by |a(4)| = √((-43)^2 + 22^2) ≈ 48.78.
The direction of acceleration at t = 4 is given by the vector (-43, 22).
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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.
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.
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.
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