Answer:
It's position at time t = 5 is 593.
Step-by-step explanation:
The velocity v(t) is the integral of the acceleration a(t)
The position s(t) is the integral of the velocity v(t)
We have that:
The acceleration is:
Velocity:
K is the initial velocity, that is v(0). Since V(0) = 13, K = 13
Then
Position:
Since s(0) = 3
What is its position at time t=5?
This is s(5).
It's position at time t = 5 is 593.
First of all, I'm going to assume that we have a concave down parabola, because the stream of water is subjected to gravity.
If we need the vertex to be at , the equation will contain a
term.
If we start with we have a parabola, concave down, with vertex at
and a maximum of 0.
So, if we add 7, we will translate the function vertically up 7 units, so that the new maximum will be
We have
Now we only have to fix the fact that this parabola doesn't land at , because our parabola is too "narrow". We can work on that by multiplying the squared parenthesis by a certain coefficient: we want
such that:
Plugging these values gets us
As you can see in the attached figure, the parabola we get satisfies all the requests.
Answer: sqrt(2)/2 which is choice D
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Explanation
(3pi/4) radians converts to 135 degrees after multiplying by the conversion factor (180/pi).
The angle 135 degrees is in quadrant 2. We subtract the angle 135 from 180 to find the reference angle
180-135 = 45
Then you can use a 45-45-90 triangle to determine that the ratio of opposite over hypotenuse is sqrt(2)/2
sine is positive in quadrant 2
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Alternatively, you can use a unit circle. Refer to the diagram below. In red, I've circled the angle 3pi/4 radians. The terminal point for this angle has a y coordinate of sqrt(2)/2
Recall that y = sin(theta).
