Giving the speed, find meeting points of two objects.

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Two tortoises named A and B must run a race. A starts with an average speed of 720 feet per hour. Young B knows she runs faster than A, and furthermore has not finished her cabbage.

When she starts, at last, she can see that A has a 70 feet lead but B's speed is 850 feet per hour. How long will it take B to catch A?

Could you just walk me through this?

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Every hour, $B$ gains $A$ a distance of $850-720=130$ feet. Since $70<130$, we know that the time it takes for $B$ to catch up to $A$ is less than an hour.

$6$ minutes is $\frac{1}{10}$ of an hour. In $6$ minutes, $B$ gains $130\cdot\frac{1}{10}=13$ feet on $A$.

$\frac{70}{13}$ is the amount of $6-$minute increments that it takes for $A$ to be caught up by $B$.

Thus, the total amount of time for $B$ to catch up to $A$ is $\frac{70}{13}\cdot 6=\boxed{\frac{420}{13}}\approx\boxed{32.3076923}$ minutes.

I may be wrong, as I am very tired and am practicing for the AMC 10. Tell me if I made a mistake lol. And if you want clarification just comment and I will add.

Max0815

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Hint. Essentially, when $B$ starts moving at $850$ feet per hour, $A$ is at point $70.$ In $t$ hours $B$ will have moved ??? and $A$ will have moved ???. You have to come up with two linear equations of $t$ (of the form $\alpha t + \beta$) and make them equal to find the time.

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In problems like this, you always find the catch up speed which is the difference between fastest and the slowest. The speed difference is $130$ feet per hour. Now all you need to find is how much time is needed to cover $70$ feet.