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.

2 Answers
Noah G
Oct 24, 2017

See below.

Explanation:

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#.

Dwight
Oct 24, 2017

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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