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