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```				   A particle moves in x-y plane. At time t=0, particle is at (1m,2m) and has vel.(4i+6j)m/s. At t=4sec, particle  reaches (6m,4m) and has velocity (2i+10j)m/s. In the given time interval, find:
a)Avge.Vel=?
b)Avge. Acceleration=?
c)From,given data, find avge.Speed=?
```

6 years ago

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```										Dear Chilukuri
Total displace ment = ( (6-1)2 + ( 4-2)2 )1/2
= √29  m
total time     = 4 sec
change in velocity = |(2i+10j) -(4i +6j) |  =√20

so avg velocity = total displacement /total time
=√29 /4  m/sec
avg acceleration      = change in velocity /time interval
= √20 /4  =√5 /2  m/sec2
since path of the particle is not know only initial and final position is known so avg speed can not be calculated with this given data

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

```
6 years ago

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