Question

add a velocity of 30ms-1eastwards to a velocity of 40ms-1northwards and calculate the magnitude of the direction of the resultant with the east

Answer 1

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Answer 2

The resultant velocity is 50 m/s 53° due east.

Given,

Velocity due east=30 m/s

Velocity due to north=40 m/s.

To find,

the magnitude of the resultant velocity and its direction due east.

Solution:

• The concept here is from the chapter "Motion in a Plane".

• In this question, there is to be vector addition used.

• Velocity due east, V1=30  î m/s

Velocity due north, V2=40 ĵ m/s.

The angle between V1 and V2 vectors is 90°.

The resultant R of two vectors A and B having angle θ between them is calculated as:

$|R|=\sqrt{A^{2}+B^{2} +2ABcosθ }$

The resultant velocity, Vnet will be:

$|Vnet|=\sqrt{V1^{2}+V2^{2} +2V1V2cos90 }\\|Vnet|=\sqrt{30^{2}+40^{2} +2X30X40cos90 }\\|Vnet|=\sqrt{900+1600+0}\\|Vnet|=\sqrt{2500}\\|Vnet|=50m/s.$

The resultant velocity is 50 m/s.

The resultant R of two vectors A and B makes an angle α with respect to vector A is calculated as:

tanα=(Bsinθ/(A+Bcosθ))

The direction of the resultant velocity due east will be:

tanα=(40sin90/(30+40cos90)

tanα=(40x1/(30+0))

tanα=40/30

tanα=4/3

α=53°.

The direction of the resultant velocity is 53° due east.

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