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