A car accelerates from rest at a stop sign at 3.0 m/s^2 up to 21.0 m/s, then immediately decelerates at 4.0 m/s^2 to a stop at the next stop sign. How long does the whole trip between the two signs take?
A1.7 s
B5.3 s
C7.0 s
D12 s
Answer & Solution
Correct answer: D. 12 s
1. Time to speed up: t1 = v / a1 = 21.0 m/s / 3.0 m/s^2 = 7.0 s.
2. Time to slow down: t2 = v / a2 = 21.0 m/s / 4.0 m/s^2 = 5.25 s.
3. Total time = t1 + t2 = 7.0 s + 5.25 s = 12.25 s, which rounds to 12 s.
_Source: OpenStax Physics (CC BY 4.0), Ch 3 "Acceleration", section 3.1 Acceleration_
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