# Velocity between two running animals

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?

• I think you should also directly post the original question because in this form, there are not necessary informations to solve the problem. – onurcanbektas Oct 10 '16 at 4:27
• 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) – Vidyanshu Mishra Oct 18 '16 at 10:44

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 $$t_{encounter}=100\text{min}$$

• こんにちは信行。あなたはすてきな答えを書きました。 – user.3710634 Oct 10 '16 at 10:26

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.

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).

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

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