Question

Two particles a and b start moving with velocities 20m /s and 30 root 2 m/s along x axis and at an angle 45 with x axis resp in xy plane from origin the relative velocity of a with respect to b?

Answer 1

Given that the two particles "a" and "b" move with velocities 20 m/s and 30√2 m/s.
So, Va = 20 m/s
and,Vb = 30√2 m/s
Given angle is θ = 45°
Relative velocity of "a" with respect to "b" = Vab = ?
Solution:
Vab = √ Va² + Vb² - 2 Va Vb cos θ
Putting the values,we get:
Vab = √ (20)² + (30√2)² - 2 (20) (30√2) (cos 45°)
Vab = √ 400 + 1800 - 1200
Vab = √ 1000
Vab = 31.6 m/s
This is the required answer. Hope it will help you. Thanks

Answer 2

A = 20 m/s i + 0 m/s j
B = 30cos45 m/s i + 30 sin 45 j = 21.213 m/s i + 21.213 m/s j
Plugging these in to the x and y equations gives:
In the x direction , (20 + 21.213 )m/s I = 41.213 m/s i
In the y direction, (0 + 21.213 )m/s j = 21.213 m/s j
Combining these two components into the vector gives a magnitude of:

=SQRT( 41.2132 + 21.2132 ) = 46.3519

at an angle given by the inverse tangent of 21.213 / 41.213, which is 27.23
degrees. So, the velocity of the B relative to the A is 46.3519, 27.23 degrees
north of east.