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
- JimLv 71 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))
- Keith ALv 61 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
- Wayne DeguManLv 71 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
- oubaasLv 71 month ago
acceleration a = (0-V)/Δt = -45/10 = -4.5
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- Anonymous1 month ago
a = Δv / Δt = (0-45) / 10 = -4.5
- oldprofLv 71 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