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One animal $A$ can run $100$ km/h and another animal $B$ can run $85$ km/h. Suppose the slower animal $B$ starts running $25$ meters ahead of the faster animal $A$ in a direction.

How can I calculate the time elapsed before the faster animal $A$ catches the slower animal $B$?

My question: It seems I should just assume constant velocity (is it the same as $a = 0$?), but the animals will have different velocity on take off compared to when they hit their top speed, so isn't it wrong to just assume $a = 0$? But how can I else solve this?

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  • $\begingroup$ I think you should also directly post the original question because in this form, there are not necessary informations to solve the problem. $\endgroup$ – onurcanbektas Oct 10 '16 at 4:27
  • $\begingroup$ this is quite confusing.The velocities given are 85 and100km/h and distance is 25 m while people are solving the answer with assuming that everything is inn same units (25.85,100) $\endgroup$ – Vidyanshu Mishra Oct 18 '16 at 10:44
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Given the question, I suppose you should assume a constant velocity for both animals. Then, we have $$ x_A(t)=0+100t\ ;\ x_B(t)=25+85t $$ Since A catches B at a point where $x_A=x_B$, we need $$ 100t=25+85t \Rightarrow 15t=25 \Rightarrow t_{encounter}=\frac{5}{3}h $$ or also $$ t_{encounter}=100\text{min} $$

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  • $\begingroup$ こんにちは信行。あなたはすてきな答えを書きました。 $\endgroup$ – user.3710634 Oct 10 '16 at 10:26
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All the textbook physics problems are not ideal when it comes to real world scenarios. That being said, this is a speed and time problem and it doesn't take acceleration into account. Your point is valid over here, because there's no single creature that can start with a speed of 100 km/h and maintain it all over the distance, but for the sake of solving the problem, you need to keep acceleration out, unless it's specifically mentioned.

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Imagine them both running Vertically on a 2D plane (Up the Y Axis).

When they catch each other will be when the Y coordinates of both animals (which we will show as lines,) are equal.

Therefore, you can solve this with simultaneous equations (or even parametric equations). Parametric equations seem to be easier.

A = (0t, 100t)

B = (0t, 25 + 85t)

1t = 1 Hour

We are looking for when the Y coordinates are equal, so: 100t = 25+85t

t = 5/3 when the Y coordinates are equal.

As t = 1 hour: Animal A reaches Animal B at the 1 Hour 40 minutes mark (100 minutes mark).

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It is simple,just see the motion of one of the animal relative to other. For ease let's see the motion of animal A with respect to B. since none of the animal is accelarating,thus relative accelaration is 0,also relative velocity is 15km/h(100-85) and relative distance is 0.025km(25 m). Now apply equation of motion,you will get; S(relative)=U(relative)time , at^2 is not involved since "a" is zero. so time taken=(0.025/15)hr

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  • $\begingroup$ i think you have taken speed incorrectly,for this question speed would have been in meter/sec(in my opinion) $\endgroup$ – Vidyanshu Mishra Oct 10 '16 at 5:36
  • $\begingroup$ i consider other answers wrong bcz most of them have used speed in km/h and displacement in metres $\endgroup$ – Vidyanshu Mishra Oct 10 '16 at 5:39
  • $\begingroup$ if the speeds are in m/s the answer will be 5/3 sec $\endgroup$ – Vidyanshu Mishra Oct 10 '16 at 5:40

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