At the end of its trip, the same subway train slows from 30.0 km/h (8.333 m/s) to a stop in 8.00 s. What is its average acceleration?
A3.75 m/s^2
B-0.417 m/s^2
C1.04 m/s^2
D-1.04 m/s^2
Answer & Solution
Correct answer: D. -1.04 m/s^2
1. Here v_0 = 8.333 m/s and v_f = 0.
2. delta v = 0 - 8.333 m/s = -8.333 m/s.
3. a = -8.333 m/s / 8.00 s = -1.04 m/s^2.
4. The minus sign shows the acceleration points opposite to the train's motion, which is why it slows down.
_Source: OpenStax Physics (CC BY 4.0), Ch 3 "Acceleration", section 3.1 Acceleration_
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