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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.

user130306
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  • You're right. The book is wrong. – Deepak Nov 05 '20 at 15:37
  • As far as the more general question of using positives vs using negatives... so long as the equation and calculations are set up in an appropriate manner, either could be used. Rather than talking about negative money gain, you could talk about positive money loss. Use whichever makes more sense to you and helps you avoid mistakes. – JMoravitz Nov 05 '20 at 15:39
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    If animals were deers you're unable to conclude anything about sheep collision... ;-) – CiaPan Nov 05 '20 at 15:39

2 Answers2

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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.

Ross Millikan
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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.

Deepak
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