Physics (read)Help please =)
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Physics (read)Help please =)

[From: ] [author: ] [Date: 11-12-08] [Hit: ]
Thus, the total distance the object traveled is 50 m + 20 m = 70 meters.......
An object accelerates uniformly from the rest to a velocity of 20 m/s in 5s. It then immediately begins to decelerate and comes to full stop after 2s.
a.Calculate the objects acceleration
b.Calculate the objects deceleration
c.Calculate the total distance the object traveled.

I just need to understand this last one..O_O

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Only the last one? Okay so then I'll assume you already know how to solve for the object's acceleration and deceleration.
The displacement is equal to (final velocity^2 - initial velocity^2)/(2*acceleration). Note that displacement does not necessarily equal distance traveled, but here we'll assume it does. So let's plug in the values.

During acceleration: (400 m^2/s^2 - 0 m^2/s^2)/(2 x 4 m/s^2) = 50 meters
During deceleration: (0 m^2/s^2 - 400 m^2/s^2)/(2 x -10 m/s^2) = 20 meters

Thus, the total distance the object traveled is 50 m + 20 m = 70 meters.

:D

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acceleration=4 m/s^2
deceleration=5 m/s^2
distance=75m
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