a particle is moving such that its position coordinates (x, y)are (2m, 3m)at time t=0, (6m, 7m)at time t=2s and (13m, 14m)at time t=5s.average velocity vector (v)from t=0to t=5s is.
Answer (1)
Student Expert
30th Apr, 2020
Hi
We know that average velocity is given by
= net displacement total time
Now,
net displacement = final position - intial position
= (13i + 14j) - (2i +3j)
= (13i - 2i) + (14j - 3j)
= 11i + 11j
average velocity = (11 i + 11j)/5
Then the magnitude of average velocity
= (121/25+ 121/25 ) ^1/2
= 11(2/5 )^1/2
Thankyou
We know that average velocity is given by
= net displacement total time
Now,
net displacement = final position - intial position
= (13i + 14j) - (2i +3j)
= (13i - 2i) + (14j - 3j)
= 11i + 11j
average velocity = (11 i + 11j)/5
Then the magnitude of average velocity
= (121/25+ 121/25 ) ^1/2
= 11(2/5 )^1/2
Thankyou
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