# A dripping water faucet steadily releases drops 1.0 s apart. As these drops fall, will the distance between them increase,decrease,or remain the same?

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As the time increases the distance between the two drops will also increase. This is because both drops are subject to the acceleration of gravity (9.8 m/s/s) and the first drop starts accelerating sooner so has reached a higher velocity at each one second increment of time.

Mathematically, the distance the drop has fallen is determined by the equation: delta y = v(i)t + 1/2 gt^2. Where delta y is the distance, v(i) is the initial velocity, g is acceleration of gravity, and t is the time. Since v(i) is zero, this reduces to: delta y = 1/2gt^2.

Now consider two drops able to fall a long distance.

at time zero the first drop leaves the faucet and begins accelerating.

at t = 1s, delta y = 4.9 m and the second drop leaves the faucet.

at t = 2s, delta y = 19.6 m for drop 1, 4.9 m for drop 2, a difference of 14.7 m.

at t = 3 s, delta y = 44.1 for drop 1, 19.6 m for drop 2, a difference of 24.5 m.

at t = 4 s, delta y = 78.4 m for drop 1, 44 m for drop 2, a difference of 34.4 m

at t = 5s, delta y = 122.5 m for drop 1, 78.4 m for drop 2, a difference of 44.1 m.

So over time the distance increases each second.