Question

To swing. Here, ttt is entered in radians. P(t) = -5\cos\left(2\pi t\right) + 5P(t)=−5cos(2πt)+5P, left parenthesis, t, right parenthesis, equals, minus, 5, cosine, left parenthesis, 2, pi, t, right parenthesis, plus, 5 What is the first time the pendulum reaches 3.5\text{ cm}3.5 cm3, point, 5, start text, space, c, m, end text from the place it was released? Round your final answer to the nearest hundredth of a second

Answer 1

0.20 seconds

Explanation:

Given the function:

For the distance traveled by a pendulum in time
`t`:

`P(t)=-5cos(2\pi t)+5`

Where
`t` is in radians.

Also, given that
`P(t)=3.5\ cm`

To find:

The time
`t` = ? to the nearest hundredth.

Solution:

Putting the values as per given statements, it can be observed that:

`3.5=-5cos(2\pi t)+5\\\Rightarrow -1.5=-5cos(2\pi t)\\\Rightarrow 0.3=cos(2\pi t)`

Taking the inverse:

`cos^(-1)(0.3) = 2\pi t`

`\Rightarrow 2\pi t=72.54^\circ`

We know that
`2\pi\ radians = 360^\circ`

Putting the value above:

`\Rightarrow 360t=72.54^\circ\\\Rightarrow t = (72.54)/(360)\\\Rightarrow t =\bold{0.20}\ seconds`

So, after 0.20 seconds, the pendulum reaches 3.5 cm from the place it was released.