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

Answer:

0.20 seconds

Step-by-step 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:

$\Rightarrow 360t=72.54^\circ\\\Rightarrow t = \dfrac{72.54}{360}\\\Rightarrow t =\bold{0.20}\ seconds$

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