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Suppose that you are in a car and that you observe a car behind you travelling at some speed and also that you know the speed at which your car is travelling. Suppose also that the distance between the cars is known, is it possible to calculate the time at which the car coming behind your car will overtake you?

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    $\begingroup$ What you want to google is called relative velocity, I pulled this from the first page of results schoolphysics.co.uk/age16-19/Mechanics/Kinematics/text/… $\endgroup$
    – spyr03
    Aug 11, 2015 at 20:56
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    $\begingroup$ do you assume constant speed for each car? $\endgroup$
    – Math-fun
    Aug 11, 2015 at 21:12
  • $\begingroup$ To add to the link provided by @spyr03, you might also want to graph the (1 dimensional) position of both cars against time, and the overtake point is when the two graphs (lines) intersect (Google "relative velocity graph" and you'll see what I mean) $\endgroup$ Aug 11, 2015 at 22:32
  • $\begingroup$ I agree with what you are talking about. However my main problem is that the distance between the cars will be changing time to time. $\endgroup$ Aug 13, 2015 at 18:18

2 Answers 2

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If the velocities are not constant then you'll need calculus. Integrating a function of velocity over a time interval gives the total distance traveled during that time interval. So, if car $A$ is initially $(t=0)$ a distance $d_0$ ahead of car $B$, and the velocities of the cars are given by $v_A(t)$ and $v_B(t)$ respectively where $v_B(t)>v_A(t)$, then we want to find the point in time $t$ where the difference in positions are $0$: $$\int_0^tv_B(t)\ dt - \left(\int_0^tv_A(t)\ dt+d_0\right)=0.$$ The above is a fancy way of stating that the total distance traveled is equal. If the velocities are constant, then just as @Harish said, \begin{align} \notag \int_0^tv_B(t)\ dt - \left(\int_0^tv_A(t)\ dt+d_0\right)&=0\\ \notag \int_0^tV_B\ dt - \left(\int_0^tV_A\ dt+d_0\right) &= 0\\ \notag V_Bt-V_At-d_0 &= 0\\ \notag t &= \frac{d_0}{V_B-V_A}. \end{align}

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  • $\begingroup$ I also accept this one again. $\endgroup$ Aug 13, 2015 at 19:11
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Nice problem

Let $V_A$ be the speed of your car A & $V_B$ be the speed of the car B behind you.

Since car B is to overtake you hence its speed is taken greater than that of your car i.e. $V_B>V_A$

Now, the relative velocity of car B approaching your car is given as $$V_B-V_A$$

Now, at any moment, let $d$ be the distance between your car A & car B behind you then the time taken by car B to overtake you is $$=\frac{\text{distance between car A & B}}{\text{relative velocity of car B w.r.t. car A}}$$ $$=\frac{d}{V_B-V_A}$$

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  • $\begingroup$ Do you think problems can ever be nice Mr.Harish? :) $\endgroup$ Aug 13, 2015 at 18:17
  • $\begingroup$ I appreciate. Its a nice answer and it makes sense. $\endgroup$ Aug 13, 2015 at 18:20
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    $\begingroup$ Sufyan Naeem, a maths problem is not a trouble. $\endgroup$ Aug 13, 2015 at 18:22

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