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

Respuesta :

Answer:

Distance between them increase

Explanation:

The position S of the water droplet can be determined  using equation of motion

[tex]S=ut+\frac{1}{2} at^2[/tex]

where [tex]u[/tex] is the initial velocity which is zero here

[tex]t[/tex] is time taken, [tex]a[/tex] is acceleration due to gravity

the position of  first drop after time [tex]t[/tex] is given by

[tex]S_{1} =0 \times t+ \frac{1}{2} at^2=\frac{1}{2} at^2............(1)[/tex]

the position of  next drop at same time is

[tex]S_{2} =\frac{1}{2} a(t-1)^2 = \frac{1}{2} a(t^2+1-2t)............(2)[/tex]

distance between them is [tex]S_{1} -S_{2}[/tex]  is [tex]a(t-1)[/tex]

from the above the difference will increase with the time

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