# The velocity of a particule is v = 2t + cos (2t). When t = k the acceleration is 0. Show that k = pi/4?

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t = time

for t, 0 < t < 2.

t = time

for t, 0 < t < 2.

See below.

The derivative of velocity is acceleration, that's to say the slope of the velocity time graph is the acceleration.

Taking the derivative of the velocity function:

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Since the acceleration is the derivative of the velocity,

The acceleration function must be

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To show that ( k = \frac{\pi}{4} ), we need to find the value of ( k ) such that the acceleration (( a )) is ( 0 ) when ( t = k ).

Given ( v = 2t + \cos(2t) ), we can find acceleration (( a )) by differentiating velocity (( v )) with respect to time (( t )).

[ a = \frac{dv}{dt} = 2 - 2\sin(2t) ]

Now, setting ( a = 0 ), we have:

[ 0 = 2 - 2\sin(2k) ]

[ \sin(2k) = 1 ]

The only solution for ( \sin(2k) = 1 ) is when ( 2k = \frac{\pi}{2} ) or ( k = \frac{\pi}{4} ). Hence, ( k = \frac{\pi}{4} ).

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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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