Question

A particle is moving such that its position coordinates (x, y) are (2 m, 3 m) at time t = 0, (6 m, 7 m) at time t = 2 s and (13 m, 14 m) at time t = 5 s. average velocity vector av (v ) from t = 0 to t = 5 s is

Answer 1

as you know, average velocity = net displacement/total time ,

net displacement = final position - intial position
= (13i + 14j) - (2i +3j)
= (13i - 2i) + (14j - 3j)
= 11i + 11j
hence, net displacement = 11 i + 11 j
average velocity = (11 i + 11j)/5

now magnitude of average velocity =
$\sqrt{\frac{121}{25}+\frac{121}{25}}$
= 11√2/5

Answer 2

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