Velocity vector $A$ has components $A_x=3 \text{ m/s}$ and $A_y=4 \text{ m/s}$. A second velocity vector $B$ has a magnitude that is twice that of $A$ and is pointed down the negative $x$-axis. Find the $x$ component of $B$
This is what I did: Magnitude of A=5 so magnitude of B is 10 Bx= 10cos180=-10 Is this correct? What I am hesitant about is this sentence "pointed down the negative $x$-axis"
The magnitude of $A$ is $|A|=\sqrt {3^2 + 4^2} = 5$ and we know that $|B| = 2|A| = 2 \cdot 5 = 10$. The part about $B$ pointing down the negative axis means $B_y = 0$ and $B_x < 0$. Using this we know, $\sqrt{B_x^2 + 0^2} = 10 \implies |B_x| = 10 $. And since the problem states it is pointing in the negative direction, $B_x = -10$.
Or another method, as you've shown is to use the angle of vector $B$ and $B_x = |B|\cos \theta$. In this example $\theta = \pi = 180^\circ$, and therefore, $B_x = 10 \cos \pi = -10$.