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 ( ) vav 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 =

= 11√2/5

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Answer 2

Answer:

The average velocity is $\dfrac{11i+11j}{5}$.

Explanation:

Given that,

Position coordinates at (t = 0) = 2m, 3m

Position coordinates at (t = 2) = 6m, 7m

Position coordinates at (t = 5) = 13m, 14m

We need to calculate the net displacement

Using formula of net displacement

$Net\ displacement=final\ position-initial\ position$

Put the value into the formula

$Net\ displacement=(13i+14j)-(2i+3j)$

$Net\ displacement =11i+11j$

We need to calculate the average velocity

Using formula of average velocity

$V_{av}=\dfrac{D}{t}$

Where, D = net displacement

t = total time

Put the value into the formula

$\vec{v_{av}}=\dfrac{11i+11j}{5}$

Hence, The average velocity is $\dfrac{11i+11j}{5}$.