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user100 [1]
3 years ago
5

el extremo de un brazo robotico se mueve hacia la derecha a 8m/s . cuatro segundos despues. se mueve hacia la izquierda a 2m/s.¿

cual es su cambio de velocidad y cual su cambio de aceleracion?
Physics
1 answer:
rewona [7]3 years ago
3 0

Answer: Change in velocity: -10 m/s, acceleration=-2.5m/s

Explanation:

The question in english is written below:

The end of a robotic arm moves to the right at 8m s. Four seconds later it moves to the left at 2m/s. What is its change in velocity and what is its change in acceleration?

We have the following data:

Initial Velocity of the robotic arm: V_{o}=8 m/s (to the right)

Final Velocity of the robotic arm: V_{f}=-2 m/s (to the left)

Variation in time: \Delta t=4 s

Firstly we have to calculate the change in velocity \Delta V, which is given by the following equation:

\Delta V=V_{f}-V_{o}=-2 m/s-8 m/s

\Delta V=-10 m/s This means the resultant velocity is to the left

Now, we can find the acceleration a, given by:

a=\frac{\Delta V}{\Delta t}

a=\frac{-10 m/s}{4 s}

a=-2.5 m/s^{2} This means the acceleration is also to the left, however its magnitude is always positive: 2.5 m/s^{2}

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Answer:

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Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

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x=V_{ox}t

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\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

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tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

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Explanation:

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Replacing 4 in 5

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