Ok, so I was wondering if I could get a clear explanation of why we solve relative motion problems the way we do?

so I'm given the problem where I have 2 cars, car A and Car B. Now car A is traveling at 24km/hr relative to the earth, and car B is traveling 13km/hr relative to the hr. Now everywhere I looked it tells me, to get the velocity of car A relative to Car B, you essentially subtract 13 from 24 and you have your answer. Now, where I'm confused is how the operation of subtracting the 2 vectors gives us the velocity of A with respect to B?. Essentially the explanation given from my teacher is, "Do steps 1,2 and 3 and you have your answer", however, i want to know why we do what we do to get the answer, if that makes sense.


Suppose you were sitting in car $A$ then the velocity of Car $B$ that you perceive will be the relative velocity of the car $B$ relative to you (or car $A$).

For example, the velocities of the cars with respect to the earth are,

$\vec{V_a}$ = 24 $\frac{km}{hr}\hat{i}\space and\space \vec{V_b}$ = 13 $\frac{km}{hr}\hat{i}$

Then to obtain the velocity of car $B$ relative to car $A$, imagine what velocity car $B$ will you perceive if you were sitting in car $A$.

In such questions, the observers think of themselves to be at rest.

$Velocity\space of\space B\space with\space respect\space to\space A = \vec{V_{ba}}$ = $\vec{V_b}-\vec{V_a}$


$\vec{V_{ba}}=11\frac{km}{hr}\hat{-i}$ (Notice the -ve sign here)

The negative sign indicates that the car $B$ (wrt to you) will come backward (as you think of yourself at rest).

This means that when sitting in the car $A$, you will perceive that the car $B$ is coming towards you with the speed $11\frac{km}{hr}$.

Now if we want to find the velocity of $A$ relative to $B$, sit in the car $B$ and the velocity of car $A$ that you now perceive will be the velocity of car $A$ relative to you.


Here, we get a positive value, which means that you (sitting in car B) will perceive the car $A$ come towards you with speed $11\frac{km}{hr}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.