Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

what is the acceleration of a plane moving at 45 m/s and come to the rest in 10 s?

units are m/s^2 because if formatting issues please do not include units in your answer this time ​

6 Answers

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  • Jim
    Lv 7
    1 month ago

    a = ∆v/∆t

    a = (0-45)/10 m/s²

    a = -4.5 m/s²  ((the negative means deceleration since it's a vector))

    a = -4.5 ((without units))

  • 1 month ago

    The plane has to lose  10 m/s  in  10 sec.

    With constant acceleration, that is

    -45/10 m/s^2 = -4.5 m/s

  • 1 month ago

    Using v = u + at we have,

    0 = 45 + 10a

    so, 10a = -45

    Hence, a = -4.5 ms⁻²

    i.e. deceleration of 4.5 ms⁻²

    :)> 

  • oubaas
    Lv 7
    1 month ago

    acceleration a = (0-V)/Δt = -45/10 = -4.5 

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  • Anonymous
    1 month ago

    a = Δv / Δt = (0-45) / 10 = -4.5

  • 1 month ago

    Actually, you can take out the units.  You really need to do units analysis when working this kind of physics problems.  So 45 m/s//10 s = 45/10 m/s//s = 4.5 m/s//s = 4.5 m/s^2

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