A girl rolls a basketball with an initial velocity of 2.0 m/s. It decelerates at a rate of -0.20 m/s^2 and rolls to a stop exactly 5.0 seconds after 12:00 p.m. At what time did she start the ball rolling?
A5 seconds after noon
B0.1 seconds before noon
C0.1 seconds after noon
D5 seconds before noon
Answer & Solution
Correct answer: D. 5 seconds before noon
1. Use v = v_0 + at with v = 0 at the stop: 0 = 2.0 m/s + (-0.20 m/s^2) t.
2. Solving gives t = 2.0 / 0.20 = 10 s of rolling before it stopped.
3. The ball stopped 5.0 s after noon, so it started rolling 10 s before that moment.
4. 5.0 s after noon minus 10 s is 5 s before noon.
_Source: OpenStax Physics (CC BY 4.0), Ch 3 "Acceleration", section 3.1 Acceleration_
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