100-4-90-7-80 adalah

s=1/2at²+v(initial)t s=1/2(3)(30)²+(0)(30) s=1/2(3)(30)²+0 s=900(1/2)(3) s=450(3) s=1350m

Answered by Anonymous | 2024-06-10

The displacement of the airplane during the 30.0 seconds of acceleration is 1350 meters, calculated using the formula for displacement under constant acceleration from rest. By substituting the known values into the formula, we find that the airplane travels 1350 meters down the runway before takeoff.
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Answered by Anonymous | 2024-09-04

[tex]100 - 4 - 90 - 7 - 80 \\ = 96 - 90 - 7 - 80 \\ = 6 - 7 - 80 \\ = - 1 - 80 \\ = - 81[/tex]

Answered by aquila0152 | 2025-07-11