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Because this is a 2-D situation, we have to write this as two separate equations, one for the x-components and one for the y-components :
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.
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.
anaik:
actually it's 30 root 2
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