The problem is as follows:
In a race track an orange car $A$ is moving to the north ($\textrm{Y-axis}$) at a rate of $20\frac{m}{s}$ with respect to that a tv cameraman who is at a point labeled $O_{1}$ in the ground. Simultaneously another car (blue) $B$ is moving to the direction known as $N53^{\circ}W$ at $25\frac{m}{s}$ with respect to that another cameraman $O_{2}$. If the second cameraman $O_{2}$ is holding a camera in a dolly moving to the same direction of the blue car $B$ and the velocity $v_{o_{2}o_{1}}$ is $5.0\,\frac{m}{s}$. Find the velocity of the blue car $B$ with respect to that $A$ in $\frac{m}{s}$.
The given alternatives on my book are:
$\begin{array}{ll} 1.&-15\hat{i}+4\hat{j}\frac{m}{s}\\ 2.&-24\hat{i}-2\hat{j}\frac{m}{s}\\ 3.&-15\hat{i}-2\hat{j}\frac{m}{s}\\ 4.&+26\hat{i}+2\hat{j}\frac{m}{s}\\ 5.&-26\hat{i}+4\hat{j}\frac{m}{s}\\ \end{array}$
Okay I'm lost with this problem. Essentially my source of confusion is how should I understand
$v_{o_{2}o_{1}}= 5.0\,\frac{m}{s}$
My only guess here is that what the author intended to explain was that the velocity of the blue car with respect to that the first cameraman is $5\frac{m}{s}$. But other than that. I don't know exactly what else should I do with the given information to work with the given bearing angles.
Can somebody help me here?.
I'll be honest, I don't think very highly of the way this question was posed to you. It seems unnecessarily disorganized and confusing.
But we have a sentence starting with, "If the second cameraman $O_2$ is holding a camera in a dolly", followed by a direction and a "velocity" which is actually a speed (specifically, $5.0 \frac{\mathrm m}{\mathrm s}$), it seems the only thing this can be describing is the speed and direction of travel of the dolly. "The same direction of the blue car $B$" is just a convoluted way of saying the direction is $N53^\circ W.$
One of the puzzling things about this problem statement is why use so many words to establish cameraman $O_1$ as a reference point and then not mention in words that the cameraman is the reference point with respect to which the velocity of the dolly was described. Instead we are left to guess this because the two subscripts of $v_{O_2O_1}$ are the identifiers of the two cameramen, $O_1$ and $O_2$; or perhaps because no reference point was named, we are supposed to interpret the $5.0 \frac{\mathrm m}{\mathrm s}$ as a speed relative to the ground, which makes it also a speed relative to $O_1$ (who doesn't move relative to the ground).
In any case, you have $O_2$ moving at $5.0 \frac{\mathrm m}{\mathrm s}$ in the direction $N53^\circ W$ relative to $O_1$ (or, equivalently, relative to the ground), you have a blue car $B$ moving $25.0 \frac{\mathrm m}{\mathrm s}$ in the direction $N53^\circ W$ relative to $O_2$, and you have an orange car $A$ moving north at $20.0 \frac{\mathrm m}{\mathrm s}$ relative to $O_1.$
Once you decide what each velocity vector is and what it signifies (what is moving at that velocity relative to what), everything is just vector addition and subtraction with a few rules. In the following, I use letters $v$ and $w$ to denote velocities (i.e., each letter stands for a speed and direction):
If D is moving at velocity $v$ relative to C, then C is moving at velocity $-v$ relative to D.
If D is moving at velocity $v$ relative to C and E is moving at velocity $w$ relative to D, then E is moving at velocity $v+w$ relative to C.
If C is moving at velocity $v$ relative to D and E is moving at velocity $w$ relative to D, then E is moving at velocity $w-v$ relative to C.
You can derive the third rule from the first two, but I still like to keep it in mind so that it is readily available to me.
Of course to actually add or subtract any of the vectors in the problem, I would convert each of the vectors to $\hat\imath$ and $\hat\jmath$ components. A picture often helps, because then for oblique directions like $N53^\circ W$ you can draw a right triangle with the legs parallel to the axes and the vector on the hypotenuse.