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. An object has a position given by ~r(t) = [3.0 m − (4.00 m/s)t]ˆı + [6.0 m − (8.00 m/s2 )t 2 ]ˆ , where all quantities are in SI units. What is the magnitude of the acceleration of the object at time t = 5.00 s? (a) 8.94 m/s 2 (b) 0.00 m/s 2 (c) 16.0 m/s 2 (d) 8.00 m/s 2 (e) 4.00 m/s 2

Respuesta :

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Answer:

(c) 16 m/s²

Explanation:

The position is [tex]r(t) = [3.0 \text{ m} - (4.00 \text{ m/s})t]\hat{i} + [6.0 \text{m} - (8.00 \text{ m/s}^2 )t^2 ]\hat{j}[/tex].

The velocity is the first time-derivative of r(t).

[tex]v(t) = \dfrac{d}{dt}r(t) = -4.00\,\hat{i} -16t\,\hat{j}[/tex]

The acceleration is the first time-derivative of the velocity.

[tex]a(t) = \dfrac{d}{dt} v(t) = -16\hat{j}[/tex]

Since a(t) does not have the variable t, it is constant. Hence, at any time,

[tex]a = -16\hat{j}[/tex]

Its magnitude is 16 m/s².

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