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kodGreya [7K]
3 years ago
14

Graham and Hunter are circus performers. A cable lifts Graham into the air at a constant speed of 1.5 ft/s. When Graham’s arms a

re 18 ft above the ground, Hunter, who is standing directly underneath Graham, throws Graham a ball as the cable continues to lift him higher. Hunter throws the ball from a position 5 ft above the ground with an initial velocity of 24 ft/s. Which system of equations can be used to model this situation?
A) h=18+1.5t
h=5+24t-16t^2

B) h=18+1.5t
h=5+24t+16t^2

C) h=18+1.5t
h=5+24t

D) h=18t+1.5t-16t^2
h=5+24t-16t^2
Mathematics
1 answer:
lidiya [134]3 years ago
5 0

Answer: Hello there!

this type of equations in one dimension (when all the factors are constants) are written as:

h =  initial position +  initial velocity*t + (acceleration/2)*t^2

First, let's describe the hunter's equation:

We know that Graham moves with a velocity of 1.5 ft/s, and when he is  18 ft above the ground, Hunter throws the ball, and because Graham is pulled with a cable, he is not affected by gravity.

If we define t= 0 when Graham is 18 ft above the ground, the equation for Graham height (in feet) is:

h = 18 + 1.5t

where t in seconds.

Now, the equation for the ball:

We know that at t= 0, the ball is thrown from an initial distance of 5ft, with an initial velocity of 24ft/s and is affected by gravity acceleration g, where g is equal to: 32.2 ft/s (notice that the gravity pulls the ball downwards, so it will have a negative sign)

the equation for the ball is:

h = 5 + 24t - (32.2/2)t^2 = 5 + 24t - 16.1t^2

So the system is:

h = 18 + 1.5t

h = 5 +24t - 16.1t^2

so the right answer is A

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Answer:

Step-by-step explanation:

From the given information:

R_{in} = ( \dfrac{1}{2} \ lb/gal) (6)\ gal /min \\ \\R_{in} = 3 \ lb/min

Given that the solution is pumped at a slower rate of 4gal/min

Then:

R_{out} = \dfrac{4A}{100+(6-4)t}

R_{out}= \dfrac{2A}{50+t}

The differential equation can be expressed as:

\dfrac{dA}{dt}+ \dfrac{2}{50+t}A = 3  \ \ \ ... (1)

Integrating the linear differential equation; we have::

\int_c \dfrac{2}{50 +t}dt = e^{2In |50+t|

\int_c \dfrac{2}{50 +t}dt = (50+t)^2

multiplying above integrating factor fields; we have:

(50 +t)^2 \dfrac{dA}{dt} + 2 (50 + t)A = 3 (50 +t)^2

\dfrac{d}{dt}\bigg [ (50 +t)^2 A \bigg ] = 3 (50 +t)^2

(50 + t)^2 A = (50 + t)^3+c

A = (50 + t) + c(50 + t)²

Using the given conditions:

A(0) = 20

⇒ 20 = 50 + c (50)⁻²

-30 = c(50) ⁻²

c = -30 × 2500

c =  -75000

A = (50+t) - 75000(50 + t)⁻²

The no. of pounds of salt in the tank after 35 minutes is:

A(35) = (50 + 35) - 75000(50 + 35)⁻²

A(35) = 85 - \dfrac{75000}{7225}

A(35) =69.6193 pounds

A(35) \simeq 70 pounds

Thus; the number of pounds of salt in the tank after 35 minutes is 70 pounds.

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