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Find the indicated velocities and accelerations.A rocket follows a path given by $y=x-\frac{1}{90} x^{3}$ (distances in miles). If the horizontal velocity is given by $v_{x}=x,$ find the magnitude and direction of the velocity when the rocket hits the ground (assume level terrain) if time is in minutes.

Calculus 1 / AB

Chapter 24

Applications of the Derivative

Section 3

Curvilinear Motion

Derivatives

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The acceleration of a rock…

So here we must find out how long the, um essentially how long the rocket is in the air. So this is you were simply going to be using vertical motion. We know that here it has no initial ah, velocity in the ex direction. And for on Lee, the why component, We're going to say that, uh, we're going to say that downwards is positive on Lee for the wide component, not for the x component. So we can say that the change and why the change in this placement in the right direction would be equal to the initial velocity of the wind direction plus t plus 1/2 a sub y t squared. We know this is going to be zero because the rocket is launched perfectly horizontally and so we can say that T would be equal to the square root of two times 30 0.0 meters. This would be divided by 9.8 meters per second squared. This is equaling 2.474 seconds. So that's how long the rock is in the air. And then when we want to find it's, uh, total horizontal, um, displacement. We can say that here. Um, we first need to find a velocity function so we can say that the velocity function in the extraction with respect itty would be equal to the initial ex velocity plus the integral from zero to t. T being what we had just found 2.474 times the acceleration in the ex direction. Because this rocket does have an acceleration in the expert in the ex direction in terms of tee times t prime and then times d t prime. So we get this in terms of tea and so we can say that this is going to be equal two 12.0 meters per second plus and then this would be 0.800 meters per second cube multiplied by t squared. We're going to integrate again in order to get the exposition. So we can then say that Delta X would be equal to 12 0.0 meters per second times t plus 0.2667 approximately meters per second cube multiplied by t cubed now T is equaling again to point 474 seconds. So Delta X is going to be equal to 12.0 meters per second multiplied by 2.474 seconds plus 0.2667 meters per second cube multiplied by 2.474 seconds. Quantity cubed and so are total displacement in the ex direction would be equal to 33 0.7 meters. This would be our final answer. That is the end of the solution. Thank you for what?

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