Question #256be

1 Answer
Nov 14, 2015

#8.70# units with an angle of #16.70^\circ#

Explanation:

The first vector is heading along the #x# axis, with a magnitude of 4 units.

The second vector has an angle of #30^{\circ}#, with a magnitude of 5 units. Decomposing this vector into its #x# and #y# components we have:

#x# component: #5cos(30)=4.33#
#y# component: #5sin(30)=2.50#

Let's call the first vector #\vec{u}# and the second #\vec{t}#.

#\vec{u}=4\hat{i}#
#\vec{s}=4.33\hat{i}+2.50\hat{j}#

The sum of the two vectors will be:

#\vec{t}=\vec{u}+\vec{s}=4\hat{i}+4.33\hat{i}+2.50\hat{j}=8.33\hat{i}+2.50\hat{j}#

The magnitude #|\vec{t}|=\sqrt{8.33^2+2.50^2}=75.64=8.70#
The angle is #\arctan(2.5/8.33)=16.70^\circ#