blackstreet23
  • blackstreet23
Suppose that the position vector for a particle is given as a function of time by vector r (t) = x(t)i + y(t)j, with x(t) = at + b and y(t) = ct2 + d, where a = 1.30 m/s, b = 1.35 m, c = 0.129 m/s2, and d = 1.14 m. (a) Calculate the average velocity during the time interval from t = 2.05 s to t = 3.95 s.
Physics
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SOLVED
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katieb
  • katieb
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blackstreet23
  • blackstreet23
@freckles @SithsAndGiggles
blackstreet23
  • blackstreet23
@Hero
blackstreet23
  • blackstreet23
@SolomonZelman

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IrishBoy123
  • IrishBoy123
\(\vec r(t) =< 1.3t + 1.35, \; 0.129t^2 + 1.14>\) \(\vec r(3.95) - \vec r(2.05) = ???\) [that is the difference in positions at these two times] \(|\vec r(3.95) - \vec r(2.05)| = ???\) [that is the actual distance apart at these two times, use Pythagoreas] \(\bar v = \dfrac{|\vec r(3.95) - \vec r(2.05)|}{3.95 - 2.05}\) [average velocity is distance travelled divided by time]

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