The position-time equation for a certain train is

Physics

1 answer:

astraxan [27]4 years ago

8 0

Answer:

$a=4.8m/s^2$

Explanation:

Hello,

In this case, since the acceleration in terms of position is defined as its second derivative:

$a=\frac{d^2x(t)}{dt^2}=\frac{d^2}{dt^2}(2.9+8.8t+2.4t^2)$

The purpose here is derive x(t) twice as follows:

$a=\frac{d^2x(t)}{dt^2}=\frac{d}{dt}(8.8+2*2.4*t)\\ \\a=4.8m/s^2$

Thus, the acceleration turns out 4.8 meters per squared seconds.

Best regards.

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THE GREEN HOSE:
Define the (x,y) coordinate at a height of 4 feet up from the ground to match where Majra is holding the green hose.
This means that the equation for the green hose is of the form
y = a(x - h)² + 4 (1)

Because water from the green hose lands on the ground 10 feet from where Majra is standing, therefore
y(10) = -4 (2)

Because the curve passes through (0,0), therefore
ah² + 4 = 0
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To satisfy (2), obtain
a(10 - h)² + 4 = -4
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Divide (3) by (4).
h²/(10-h)² = 1/2
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h² + 20h - 100 = 0

Solve with the quadratic formula.
x = 0.5[-20 +/- √(8400)] = 4.142, - 24.142
Reject the negative solution.
The vertex is at (4.142, 4).

From (3), obtain
a = -4/4.142² = -0.2332

The equation for the green hose is
y = 0.2332(x - 4.142)² + 4

THE RED HOSE
The red hose has a vertex at (3,7), according to the equation y = -(x-3)² + 7.

A graph of y(x) for both hoses is shown in the attached figure.

Answers:
a. The red hose will throw the water higher.

b. The equation for the green hose is
y = -0.2332(x - 4.124)² + 4,
with the origin at a height of 4 feet above ground level.

c. The domain for the green hose that makes sense is 0 ≤ x ≤ 10 feet.
The corresponding range is -4 ≤ y ≤ 4 feet.