Let \( t \) be the time in seconds after both vats have started being filled.
For the first vat, which starts with 14 gallons and is being filled at a rate of 20 gallons per second, the amount of liquid in the first vat after \( t \) seconds can be expressed as:
\[
\text{Amount in first vat} = 14 + 20t
\]
For the second vat, which starts with 35 gallons and is being filled at a rate of 16 gallons per second, the amount of liquid in the second vat after \( t \) seconds can be expressed as:
\[
\text{Amount in second vat} = 35 + 16t
\]
We want to find \( t \) such that both vats have the same amount of liquid:
\[
14 + 20t = 35 + 16t
\]
To solve for \( t \), we first will simplify the equation by subtracting \( 16t \) from both sides:
\[
14 + 20t - 16t = 35
\]
This simplifies to:
\[
14 + 4t = 35
\]
Next, subtract 14 from both sides:
\[
4t = 21
\]
Now, divide both sides by 4:
\[
t = \frac{21}{4} = 5.25
\]
Thus, after \( t = 5.25 \) seconds, both vats will have the same amount of liquid.
To verify, we can calculate the amounts in both vats after 5.25 seconds:
1. First vat:
\[
14 + 20(5.25) = 14 + 105 = 119 \text{ gallons}
\]
2. Second vat:
\[
35 + 16(5.25) = 35 + 84 = 119 \text{ gallons}
\]
Since both vats have 119 gallons, the solution is confirmed.
The final answer is:
\[
\boxed{5.25}
\] seconds.