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tino4ka555 [31]
3 years ago
15

The sum of two rational numbers will always be

Mathematics
2 answers:
telo118 [61]3 years ago
6 0
It's the same as adding 2 fractions the product of 2 rational numbers is rational
4vir4ik [10]3 years ago
6 0
The sum of two rational numbers will always be Rational.
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Jamie is riding a Ferris wheel that takes fifteen seconds for each complete revolution. The diameter of the wheel is 10 meters a
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Answer:

The answers to the question is

(a) Jamie is gaining altitude at 1.676 m/s

(b) Jamie rising most rapidly at t = 15 s

At a rate of 2.094 m/s.

Step-by-step explanation:

(a) The time to make one complete revolution = period T = 15 seconds

Here will be required to develop the periodic motion equation thus

One complete revolution = 2π,

therefore the  we have T = 2π/k = 15

Therefore k = 2π/15

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also we note that the center of the wheel is 6 m above ground

We write our equation in the form

y = 5*sin(\frac{2*\pi*t}{15} )+6

When Jamie is 9 meters above the ground and rising we have

9 = 5*sin(\frac{2*\pi*t}{15} )+6 or 3/5 = sin(\frac{2*\pi*t}{15} ) = 0.6

which gives sin⁻¹(0.6) = 0.643 =\frac{2*\pi*t}{15}

from where t = 1.536 s

Therefore Jamie is gaining altitude at

\frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi t}{15}) = 1.676 m/s.

(b) Jamie is rising most rapidly when   the velocity curve is at the highest point, that is where the slope is zero

Therefore we differentiate the equation for the velocity again to get

\frac{d^2y}{dx^2} = -5*(\frac{\pi *2}{15} )^2*sin(\frac{2\pi t}{15}) =0, π, 2π

Therefore -sin(\frac{2\pi t}{15} ) = 0 whereby t = 0 or

\frac{2\pi t}{15} = π and t =  7.5 s, at 2·π t = 15 s

Plugging the value of t into the velocity equation we have

\frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi t}{15}) = - 2/3π m/s which is decreasing

so we try at t = 15 s and we have \frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi *15}{15}) = \frac{2}{3} \pim/s

Hence Jamie is rising most rapidly at t = 15 s

The maximum rate of Jamie's rise is 2/3π m/s or 2.094 m/s.

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