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 #?

Answer 1

#(-43,22)#

Acceleration at #t = 4# is
#a_x(4) = d/dt(v_x) = 5 - 3t^2 = 5 - (3 × 4^2) = 5 - 48 = -43#
#a_y(4) = d/dt(v_y) = 6t - 2 = (6 × 4) - 2 = 22#
#a = (-43, 22)#

Direction:

#Tan θ = a_y/a_x = 22/(-43)#
#θ = Tan^-1(-22/43)#
Where #theta# is angle with #x#-axis
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Answer 2

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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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.

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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