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kirill
2 months ago
9

Antonio's toy boat is bobbing in the water under a dock. The vertical distance HHH (in \text{cm}cmstart text, c, m, end text) be

tween the dock and the top of the boat's mast ttt seconds after its first peak is modeled by the following function. Here, ttt is entered in radians. H(t) = {5}\cos\left({\dfrac{2\pi}{3}}t\right) - {35.5}H(t)=5cos( 3 2π ​ t)−35.5H, left parenthesis, t, right parenthesis, equals, 5, cosine, left parenthesis, start fraction, 2, pi, divided by, 3, end fraction, t, right parenthesis, minus, 35, point, 5 How long does it take the toy boat to bob down from its peak to a height of -35\text{ cm}−35 cmminus, 35, start text, space, c, m, end text? Round your final answer to the nearest tenth of a second.
Mathematics
2 answers:
Inessa [12.5K]2 months ago
4 0

Response:

0.7 seconds

Detailed explanation:

Zina [12.3K]2 months ago
3 0

Answer: The elapsed time t equals 33.0 seconds.

Detailed explanation:

We start with the equation modeling the vertical distance H between the dock and the boat's mast at time t seconds after the first peak:

H(t) = 5cos( 2π/3 ​t) − 35.5H

The maximum height is 5 units.

When the mast is at its lowest point, H(t) equals 0.

5cos( 2π/3 ​t) − (35.5/100)H = 0

5cos( 2π/3 ​t) − 0.355 × 5 = 0

5cos( 2π/3 ​t) − 0.1775 = 0

Thus, 5cos( 2π/3 ​t) = 0.1775

From there, we find: cos( 2π/3 ​t) = 0.1775/5

Which simplifies to cos( 2π/3 ​t) = 0.355

To solve for t, we compute: 2π/3 ​t = cos^{-1}(0.355)

This yields: 2π/3 ​t = 69.2

Therefore, by multiplying both sides by 3, we have: 2πt = 69.2 × 3

Which results in 2πt = 207.6

Finally, solving for t gives: t = 207.6/2π

This results in t = 33.0 seconds.

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