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.

Answer 1

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:

#v' = 2 - 2sin(2t)#
We can replace #v'# by #a#.
#a = 2 - 2sin(2t)#
Now set #a# to #0#.
#0 = 2 - 2sin(2t)#
#-2 = -2sin(2t)#
#1 = sin(2t)#
#pi/2 = 2t#
#t = pi/4#
Since we know that #0 < t < 2# and the periodicity of the #sin(2x)# function is #pi#, we can see that #t =pi/4# is the only time when the acceleration will be #0#.
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Answer 2

Since the acceleration is the derivative of the velocity,

#a=(dv)/dt#
So, based on the velocity function #v(t) = 2t+cos(2t)#

The acceleration function must be

#a(t)=2-2sin(2t)#
At time #t=k#, the accelertaion is zero, so the above equation becomes
#0=2-2sin(2k)#
Which gives #2sin(2k) = 2# or #sin(2k)=1#
The sine function equal +1 when its argument is #pi/2#
So, we have #2k=pi/2# resulting in #k=pi/4# as required.
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Answer 3

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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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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