# 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/ towards north-west

2
by jafiya

2014-08-29T13:36:44+05:30
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)
2014-08-29T14:34:36+05:30

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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.

acceleration = resultant velocity divided by time
not resultant velocity - but the change of velocity vector/time duration
to explain it, there were many steps. Other wise its just 4 lines only. ask me what u do not understand here
hello. have u understood the answer ? if any doubt, mention it here.
thanks for selecting as best answer. u r welcome.