Question on constant velocity (easy)

Question

Two animals are charging at each other. They stand 20 m apart. The smaller animal charges at 6.5 m/s. The larger animal was measured to run 24 m in 3 s. At what position will the two sheep collide?

So far I have this:

$v_L = 8 m/s$, $v_s=-6.5 m/s$

$x_{0_L}= 0$, $x_{0_s}=20$

$x_L=8t$, $x_s=-6.5t+20$

I can set $x_L$ and $x_s$ equal to each other and solve. I know how to do this.

However I am confused because the solution sets these two equations equal:

$$8t=6.5t+20$$

instead of $$8t=-6.5t+20$$

Why isn't the velocity of one of the animals opposite of the other? I thought they would be opposite since they are running towards each other.

Answer 1

It is a typo. The one with the negative sign is correct. The solution should have $t=\frac{20}{14.5}$ instead of $\frac {20}{1.5}$. That would indicate which the solution is using, whether or not the negative sign is shown.

Answer 2

You're right and the book is wrong.

A much more direct approach. You can treat the scenario as if it were a single body moving at $6.5 + 8 = 14.5$ m/s (closing speed) travelling $20$m. The time taken for that is $\frac{40}{29}$ seconds, which means at the meeting point the slower sheep would have travelled $\frac{260}{29}$m, while the faster sheep would have travelled $\frac{320}{29}$m. As expected, these distances sum to $20$m.

You know the provided solution is absurd because if you solve the equation they gave, the slower sheep would have gone much further than the initial $20$m spacing at the supposed meeting point.