Question

A particle moves towards east with velocity 5m/s.
Aftet 10 seconds its direction changes towards
north with same velocity. The average
acceleration of the particle is :-
along with direction​

Answer 1

initial velocity towards east 5m/s

= 5i

final same magnitude velocity towards north = 5j

Change occurs in 10 s

Acceleration = change in velocity/time

= 5j- 5i )/ 10

= 0.5(j - i)

Magnitude = 0.5 √2 = 0.5×1.4 = 0.7

direction along unit vector = 0.5( j-i)/0.7

= 0.7( j-i)

Answer 2

Answer:

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.