find-the-indicated-velocities-and-accelerations-a-person-on-a-hoverboard-is-riding-up-a-ramp-and-fol

Question

Find the indicated velocities and accelerations.
A person on a hoverboard is riding up a ramp and follows a path described by $y=0.15 x^{1.2} .$ If $v_{x}$ is constant at $0.50 \mathrm{m} / \mathrm{s},$ find $v_{y}$ when $x=8$.

Averell Hause · Accepted Answer

here. We&#x27;re going to apply equation to 11 and here we can say that Ah, the velocity of the wind direction is equaling zero at Why equaling? Why, Max? So we can then say that zero equaling the initial velocity of 12 meters per second plus negative 2.0 meters per second squared, multiplied by T tease, then giving us 6.0 seconds. So to find the velocity in the X direction at 6.0 seconds, this would be equaling the initial X velocity of eight meters per second. Plus the acceleration in the ex direction 4.0 meters per second squared multiplied by 6.0 seconds. And this is giving us 32 meters per second. So we can then say that at why equaling? Why, Max, we can say that the velocity and the ex direction in this case, the velocity vector would be equaling 32 meters per second I hat. So we should actually just say that this would be our final answer. That is the end of the solution. Thank you for watching

Kaylee Berzsenyi · Answer

Okay. So for number six were given a velocity of a helicopter which is equal to 0.2 t squared plus tea with the time interval from 1 to 6. And we want to find the displacement. And the displacement in this case is of our velocity is equal to D X, So velocity is equal to DX over DT. So, in order to solve for the displacement we can solve for DX So we have DX equals 0.2 tea squared plus t dt. And that means we just need to set up our integral. So from time interval 1 to 6, we can just plug in r d x so zero point zero to t squared plus t Katie. And now we just solved to find our total displacement. So we&#x27;re gonna take the anti derivative, which is going to get us 0.2 Add one to T said he cute divide by three plus T squared, divided by two from 1 to 6. So now we just plug in. So we&#x27;re gonna have zero 0.2 times six cubed, which is 2 16 divided by three plus six squared is 36 divided by two. And then we&#x27;re going to subtract the second part and plug in one. So 0.2 terms one cubed is one divided by three plus one squared this one divided by two. Okay, And now we can just simplify this equation, so we&#x27;re gonna get to 16. Divided by three gives a 72 72 times 0.2 plus 36 divided by two is 18 minus point 02 divided by three is going to give us 0.666 Repeating point 00666 repeating. And then plus 1/2 which is 0.5. So that&#x27;s gonna give us 0.5 zero 666 Repeating weaken just round two. Just 0.506 Now 72 times 720.2 0.2 is going to give us 1.14 and one point 44 Not 1.141 point 44 1.44 plus 18 gives us 19.44 So this is equal to 19 0.44 minus point five 06 19.44 minus 0.506 is going to give us our final answer of 18 0.93 and that&#x27;s gonna be our displacement and meters, and that is your final answer.

Zulfiqar Ali · Answer

will be no date X physical to 80 meter and bicycle to 60 meter. You&#x27;re re X, which has been lost. Your long XX is is equal to X divided by t and be by is equal to Dean off 25 times B X, which is equal to Dan of 25. Multiply by X, divided by t and why is equal to re y. T minus 0.5 80 square and further Why is equal appeared gain 25 x minus is either 0.5 80 square. I know we can calculate T well. T is equal to four point for life. Second on the X is equal to a 17.9 meters per second, and we buy is equal to hurting point or eight meter per second and be, which is the result of been lost musical 22 point for one meter per second.

Keshav Singh · Answer

for this problem on the topic of two dimensional kinda Matics, we have a skateboarder that is traveling on a horizontal surface and traveling with an initial velocity of 3.8 m per second towards the south with a constant acceleration of 2.2 m per square second toward the east. If we let X be eastward and yb northward And the skateboarder starting at the origin T. is equal to zero. We want to know her X and Y positions after the 0.8 seconds and her X and Y velocity components At the same time of 0.8 seconds. So firstly we can use cosmetics and we know her X position is you X times T plus a half times the acceleration in the X direction, A X times a times T squared. And so this is Initial speed in the x direction, 3.8 m/s times a time of 0.8 seconds plus zero. Since she has no acceleration in the X. Direction. And so the X. Coordinator of her position is three m. We can use the same equation and find the Y coordinate of position. Why is equal to U. Y. Times T plus a half a Y times T squared. Initial velocity in the y direction is zero plus a half Times. Her acceleration in the Y direction is 2.2 m/km2 Time, the time of 0.8 seconds squared, This gives us a wide position after 0.8 seconds To be 0.7 meters. Now we want to calculate the X. And Y components of velocity after the same time. So we&#x27;ll use a new equation. V. X is equal to you, X plus X times T. And her acceleration in the X. Direction is zero. So her final velocity in the X direction is the same as a as an initial velocity in the X. Direction. So this is 3.8 m/s and this is to the east. We can use the same equation for Hawaii velocity and V. Y. Is equal to you I plus A. Y. Times T. Now her initial velocity is zero, And her acceleration is 2.2 m/km2 Times a time of 0.8 seconds, which gives her final wind velocity after zero 0.8 seconds To be 1.8 m/s. And this is to the north.

Patrick Connors · Answer

Okay, so this problem is dealing with motion graphing and it says we have an object that&#x27;s traveling from a position of negative 16 to a position of 32 meters in a span of 18 seconds, and it wants to know what the velocity of that is as well seeing the motion graph. So since I have both a positive and a negative position here, we need to have a positive and negative axes with 32 meters at our height here. Because this is my ex value. This is my position. You can think of this like a number line, and we&#x27;re starting at a negative 16 meters Now. It tells us that it goes from one extreme to the next in 18 seconds. So my X axis is always the time in seconds here. And if I&#x27;m graphing it, we start at negative 16. We end at positive 32. And because this is a constant velocity, there&#x27;s no acceleration. It&#x27;s going to look like a straight line on my position versus time graph. So that is what the position versus time graph looks like. Now. To find the actual velocity of this, we can use the slope of this line because when we&#x27;re dealing with motion graphing, the slope of a position versus time graph is equal to the velocity of our object. So in this case, the velocity is equal to the rise. So going from negative 16 to 32 is a difference of 30 to our final position, minus the negative 16. Our initial position divided by 18 seconds, 32 minus on negative 16 becomes 48 we end up with a velocity here of 2.67 meters per second.

Problem

Find the indicated velocities and accelerations. …

Problem 22 Easy Difficulty

Find the indicated velocities and accelerations.
A person on a hoverboard is riding up a ramp and follows a path described by $y=0.15 x^{1.2} .$ If $v_{x}$ is constant at $0.50 \mathrm{m} / \mathrm{s},$ find $v_{y}$ when $x=8$.

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

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Basic Technical Mathematics with Calculus

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Grace H.

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus