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7 days ago

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An object has a position given by r⃗ = [2.0 m + (5.00 m/s)t] i^ + [3.0 m − (3.00 m/s2)t2] j^ , where quantities are in SI units.

What is the speed of the object at time t = 2.00 s?

1 answer:

Zina [11.9K]7 days ago

8 0

Answer:Speed = 13.0 m/s

Step-by-step explanation:

Considering velocity, v = (ai ± bj)m/s.

Since speed is a scalar and has only magnitude, we can derive it from the velocity by calculating just the magnitude represented as:

...Equation 1

$Speed = (\sqrt{a^{2} + b^{2} } )m/s$ From the context, the position vector is denoted as

r

To determine the velocity, we differentiate r with respect to

time, t

We know that differentiating a constant results in 0.

$v = \frac{dr}{dt} = \frac{d}{dt}[(2.0+5.00t)i + (3.0 - 3.00t^{2})j] \\\\v = 0 + 5.00i + 0 - (6.00t)j$

At time t = 2.00s,

$v = 5.00i - (6.00*2.00)j = 5.00i - 12.00j$

Using equation 1, we find the speed at this moment to be 13.00m/s

$Speed = \sqrt{(5.00)^{2} + (12.00)^{2} } = \sqrt{25 + 144} \\ \\Speed = \sqrt{169} = 13.00m/s$

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1 month ago

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Explicación paso a paso:

El enunciado no está completo. El texto completo es: "Un globo se desplaza entre las ciudades A y B, que se encuentran a 1.500 m de distancia. Los ocupantes del globo observan la ciudad A con un ángulo de depresión de 27°, mientras que, para la ciudad B, el ángulo es de 36°. ¿Cuál es la altura aproximada del globo con respecto al suelo?"

El diagrama que ilustra esta situación está en el archivo adjunto. Para calcular la altura del globo, se pueden emplear las funciones trigonométricas, siendo recomendable hacer uso de la función tangente para los ángulos de depresión mencionados:

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