Question

The given vector represents a hot air balloon that experiences a 0.25m/s^2 up and 0.25 m/s^2 to the right at 45 degrees. Find the magnitude of the resultant

Answer 1

when two vectors are inclined at some angle to each other then the magnitude of resultant of two vectors is given as

$R = \sqrt{A^2 + B^2 + 2AB cos\theta}$

here A and B are the magnitude of two vectors and theta is the angle between them

now here we know that two vectors are of magnitude 0.25 m/s^2 each and they are inclined at angle of 45 degree

now we will plug in all data above

A = B = 0.25

$\theta = 45$

$a_{net} = \sqrt{0.25^2 + 0.25^2 + 2*(0.25)(0.25)cos45}$

$a_{net} = 0.46 m/s^2$

So the magnitude of resultant acceleration will be 0.46 m/s^2