# A stone A is dropped from rest down a well

A stone A is dropped from rest down a well, and in 1 s another stone B is dropped from rest. Determine the distance between the stones another second later.

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#### Solution:

We will calculate the distance the first stone travels in 2 seconds, and we will calculate the distance the second stone traveled in 1 second. We will then subtract the difference to figure out the distance between them. To do so, we will use the following equation:

$s_2=s_1+v_1t+\dfrac{1}{2}a_ct^2$

(Where $s_2$ is final displacement, $s_1$ is initial displacement, $v_1$ is initial velocity, $t$ is time, and $a_c$ is constant acceleration)

From the question, we know the following:

$s_1=0$ ft

$v_1=0$ ft/s

$a_c=32.2$ ft/$s^2$

$t=2$ s (For the first stone)

$t=1$ s (For the second stone)

Let us use our equation to calculate the distance the first stone traveled.

$s_2=s_1+v_1t+\dfrac{1}{2}a_ct^2$

$s_2=0+0+\dfrac{1}{2}(32.2)(2^2)$

$s_2=64.4$ ft

Now, we will calculate the distance the second stone traveled.

$s_2=s_1+v_1t+\dfrac{1}{2}a_ct^2$

$s_2=0+0+\dfrac{1}{2}(32.2)(1^2)$

$s_2=16.1$ ft

We can now calculate the difference between the two:

$\Delta s=64.4-16.1=48.3$ ft