An object falls from rest for 2.0 s, at which point it is still 60 m above the ground. Which order of kinematic equations correctly finds its velocity when it hits the ground?
AFirst v = v_0 + at, then v^2 = v_0^2 + 2a(d - d_0)
BFirst v^2 = v_0^2 + 2a(d - d_0), then v = v_0 + at
COnly v = v_0 + at is needed, applied once
DOnly v^2 = v_0^2 + 2a(d - d_0) is needed, applied once
Answer & Solution
Correct answer: A. First v = v_0 + at, then v^2 = v_0^2 + 2a(d - d_0)
1. First use v = v_0 + at with the known 2.0 s to get the velocity at that moment.
2. That step needs no distance value, since only time and acceleration are known so far.
3. Then use v^2 = v_0^2 + 2a(d - d_0) with the remaining 60 m to reach the ground.
4. That second equation needs a starting velocity, which is exactly what the first equation supplied.
_Source: OpenStax Physics (CC BY 4.0), Ch 3 "Acceleration", section 3.2 Representing Acceleration with Equations and Graphs_
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