Question

A particle is moving east-wards with a velocity of 5 m/sec. In 10 sec the velocity changes to 5 m/sec north-wards. The average acceleration in this time is :
( answer should come 1/√2 m/$s^{2}$ towards north-west

Answer 1

intial velocity=v1= 5m/seast wards                                   ^||^
then after time t=10sec                                                     || towards north 5m/s
v2 =5m/s north wards                                     .⇒⇒⇒⇒⇒⇒|
                                                                towards east 5m/s
Vresultant = √v1²+v2²
                =√25+25
                 =5 √2

acceleration a=velocity/time
                     = 5√2/10 
                     =√2/2 × √2/√2
                     =1/√2 towards n-w
(direction u can understand by seeing the diagram)             

Answer 2

             North (y direction)
                    ^
                    |
                    |
West <===========> east  (x direction)

Initial velocity = 5 m/sec  i                   i = unit vector along x
Final velocity = 5 m/sec  j                    j = unit vector along y

Change of velocity =  final - initial = 5 j - 5 i m/sec = 5 (j-i) m/sec

Resultant vector of the two vectors: j, and -i is =
[tex] Magnitude = \sqrt{ 5^2 + 5^2 } = 5 \sqrt{2} m/sec

Acceleration = change of velocity / time duration
       = 5 (j - i) /10 m/sec²   = 1/2 * (j - i) m/sec² .

It s magnitude = √(1²+(-1)²) / 2 = √2/2 = 1 / √2 m/sec²

Its direction is given by tan Ф = 1 * sin 90 / [ 1 - 1 cos 90 ] = 1
             So Ф = 45 deg.
It is 45 deg in counter clock wise direction from y axis.  It means North-West direction.

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I am sorry for the delay. there were interruption here - during this question.