nilzhesosa nilzhesosa
  • 24-02-2020
  • Physics
contestada

What must be the acceleration of a train in order for it to stop from 12m/s in a distance of 541m
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beingteenowfmao
beingteenowfmao beingteenowfmao
  • 24-02-2020

Answer:

x = 541 m

Dt = ?

vi = 12 m/s

vf = 0 (stop)

a = ?

Use vf^2 = vi^2 + 2a(Dx): 0^2 = (12 m/s)^2 + 2a(541 m), a = -144 m^2/s^2, a = -144 m^2/s^2/(1082 m) = -.1331 m/s/s or -.13 m/s/s

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adegokeganiyu435 adegokeganiyu435
  • 24-02-2020

Answer: aceeleration =0.133m/s^2

Explanation:

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