The equation given in the problem is:

$d = v_0 t + \frac{1}{2} a t^2$

where:

• $v_0$ is the initial velocity,
• $t$ is the time,
• $a$ is the acceleration,
• $d$ is the displacement.

Let's analyze each choice to check which one fits the given conditions:

First option:

• $v_0 = 0 \, \text{m/s}$
• $t = 5 \, \text{seconds}$
• $a = -10 \, \text{m/s}^2$

Substitute these into the equation:

$d = (0)(5) + \frac{1}{2}(-10)(5^2)$ $d = 0 - \frac{1}{2}(10)(25) = -125 \, \text{m}$

This calculation does not match the given $d = 125 \, \text{m}$, so this option is incorrect.

Second option:

• $v_0 = 30 \, \text{m/s}$
• $t = 5 \, \text{seconds}$
• $a = -10 \, \text{m/s}^2$

Substitute these values:

$d = (30)(5) + \frac{1}{2}(-10)(5^2)$ $d = 150 - \frac{1}{2}(10)(25) = 150 - 125 = 25 \, \text{m}$

This matches the displacement of $d = 25 \, \text{m}$, so this option is correct.

Third option:

• $v_0 = -30 \, \text{m/s}$
• $t = 5 \, \text{seconds}$
• $a = -10 \, \text{m/s}^2$

Substitute these values:

$d = (-30)(5) + \frac{1}{2}(-10)(5^2)$ $d = -150 - 125 = -275 \, \text{m}$

This matches the given displacement of $d = -275 \, \text{m}$, but since the initial velocity is negative and the displacement is large and negative, this seems more consistent with an object moving downward (but not matching the likely scenario described).

Conclusion:

The correct answer is the second option:

• $v_0 = 30 \, \text{m/s}$
• $t = 5 \, \text{seconds}$
• $a = -10 \, \text{m/s}^2$
• $d = 25 \, \text{m}$

Would you like further details or have any questions about this?

Here are 5 related questions to expand your understanding:

1. What happens to the displacement if the acceleration is positive instead of negative?
2. How would the displacement change if the initial velocity $v_0$ were doubled?
3. What is the significance of the negative sign in acceleration?
4. How would you determine the final velocity of the bottle after 5 seconds?
5. What other kinematic equations could be used to solve for the displacement?

Tip: When acceleration is constant, use the appropriate kinematic equations to avoid confusion between velocity, time, and displacement values.