**AP Physics 1 Question 56: Answer and Explanation**

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**Question: 56**

**6.** A bubble in a glass of water releases from rest at the bottom of the glass and rises at acceleration a to the surface in t seconds. How much farther does the bubble travel in its last second than in its first second?

- A. at
- B. (t - 1)a
- C. (t + 1)a
- D. at

**Correct Answer:** B

**Explanation:**

B The distance travelled in the first second can be found using the Big Five,

d_{first} = v_{0}t_{first} +

Using t_{first} = 1 s and v_{0} = 0 m/s since the bubble starts at rest, the equation becomes

The distance traveled in the last second can be found using another Big Five equation.

Using t_{last} = 1 s and v_{final} = v_{0} + at_{total} = at_{total}, the equation becomes

d_{last} = at_{total} – a

The question is asking for the difference in the distance, so

d_{last} – d_{first} = at_{total} – a – a = at_{total} – a = (t – 1)a