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?
 A: 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
$$
t_{encounter}=100\text{min}
$$
A: 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.
A: 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).
A: 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
