# Homework Help: Applying Relative Motion to One Dimensional Motion Equations

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1. Jul 9, 2015

### Better WOrld

<<Moderator note: LaTeX corrected>>

Problem:
> Two cars A and B move with velocity $60 kmh^{-1}$ and $70 kmh^{-1}$. After a certain time, the two cars are 2.5 km apart. At that time, car B starts decelerating at the rate 20 kmh-2. How long does Car A take to catch up with Car B?

I tried to apply Relative Motion Concept to try and solve this problem. However, I cannot understand how to apply it to this problem.

My attempt:
I tried to apply the Relative Motion Concept to this problem as follows.

$$u_{AB}=u_A-u_B=60-70=-10kmh^{-1}$$
As per the question, the separation between the two cars ie $S_{AB}=-2.5km$ after 15 minutes.

Now, since Car A catches up with Car B eventually, thus $S_{AB_{final}}=0$ and $S_{AB_{initial}}=-2.5km$
$$\Longrightarrow S_{AB_{final}}-S_{AB_{initial}}=0-(-2.5)=u_{AB}\times t + \dfrac{1}{2}a_{AB}\times t^2$$
Now $a_{AB}=a_A-a_B=0-(-20)=20kmh^{-2}$ and $u_{AB}=10kmh^{-1}$
$$\Longrightarrow 2.5=10t+10t^2$$
However, on solving this quadratic, I get a value of time which is incorrect. Would somebody please be so kind as to show me how to correctly apply the concept of Relative Motion here? I would be truly grateful for any assistance. Many thanks in advance!

Last edited by a moderator: Jul 9, 2015
2. Jul 9, 2015

### Orodruin

Staff Emeritus
You have used the wrong value for the velocity, you found earlier that $u_{AB} = -10$ km/h.