Question

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.

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`.

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.