# An object's two dimensional velocity is given by #v(t) = ( t^2, t-t^2sin(pi/3)t)#. What is the object's rate and direction of acceleration at #t=4 #?

Please see the explanation.

The magnitude of the acceleration is

I used WolframAlpha to evaluate the above at t = 4:

The direction is along the unit vector:

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To find the acceleration vector, differentiate the velocity vector with respect to time. Then evaluate the resulting expression at t = 4.

v(t) = (t^2, t - t^2sin(π/3)t)

Differentiate each component with respect to time:

a(t) = (2t, 1 - 2tsin(π/3) - t^2cos(π/3))

Evaluate at t = 4:

a(4) = (8, 1 - 8sin(π/3) - 16cos(π/3))

Now, compute the magnitude of the acceleration vector:

|a(4)| = sqrt(8^2 + (1 - 8sin(π/3) - 16cos(π/3))^2)

Finally, find the direction of acceleration by calculating the angle between the acceleration vector and the positive x-axis using:

θ = arctan((1 - 8sin(π/3) - 16cos(π/3)) / 8)

Substitute the values for sin(π/3) and cos(π/3) and compute the angle.

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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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