How do you calculate the resultant of the following vectors 4.0m/s and 10m/s with an angle of 90degrees between them?

Question

Charlotte Aaron · Accepted Answer

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Suppose <mathjax>#vec a=4 m/s#</mathjax> and <mathjax>#vec b=10 m/s#
So,resultant <mathjax>#vec c = vec a +vec b#
So, <mathjax>#|vec c| = sqrt(a^2+b^2+2ab cos theta)#</mathjax> (where, <mathjax>#theta#</mathjax> is the angle between <mathjax>#vec a#</mathjax> and <mathjax>#vec b#</mathjax>)
Thus from the above formula we get, magnitude of the resultant vector is <mathjax>#sqrt(4^2+10^2+2*410 cos 90)=10.77 m/s#
Making an angle of <mathjax>#tan ^-1(4/10)=21.80#</mathjax> degrees w.r.t <mathjax>#vec b#

Amelia Adams · Answer

undefined To calculate the resultant of two vectors with a known magnitude and angle between them, you can use vector addition. Given the magnitudes of the vectors and the angle between them:

1. Square each vector's magnitude.
2. Add the squared magnitudes.
3. Add the product of the magnitudes and the cosine of the angle between them.
4. Take the square root of the result.

Mathematically:

$$ R = \sqrt{(4.0 \, m/s)^2 + (10 \, m/s)^2 + 2 \times (4.0 \, m/s) \times (10 \, m/s) \times \cos(90^\circ)} $$

$$ R = \sqrt{(16 \, m^2/s^2) + (100 \, m^2/s^2) + 2 \times (4.0 \, m/s) \times (10 \, m/s) \times 0} $$

$$ R = \sqrt{16 + 100 + 0} \, m/s $$

$$ R = \sqrt{116} \, m/s $$

$$ R \approx 10.77 \, m/s $$

So, the resultant velocity is approximately 10.77 m/s.