Question

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.

Answer 1

@freckles
@SithsAndGiggles

Answer 2

@Hero

Answer 3

@SolomonZelman

Answer 4

\(\vec r(t) =< 1.3t + 1.35, \; 0.129t^2 + 1.14>\)

\(\vec r(3.95) - \vec r(2.05) = ???\) [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]