- #1

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My method:

deltaS= (Vf^2-Vo^2)/2a

900= (Vf^2/.4) + (Vf^2)

900= 2.5Vf^2 + Vf^2

Vf= 16 m/s

and then I used: Vf = Vo + at

16= .2t

t=80 s

0= 16 - .5t

t=32 s

total t= 112 seconds

Can anyone please confirm this?

- Thread starter miscellaneous
- Start date

- #1

- 10

- 0

My method:

deltaS= (Vf^2-Vo^2)/2a

900= (Vf^2/.4) + (Vf^2)

900= 2.5Vf^2 + Vf^2

Vf= 16 m/s

and then I used: Vf = Vo + at

16= .2t

t=80 s

0= 16 - .5t

t=32 s

total t= 112 seconds

Can anyone please confirm this?

- #2

Doc Al

Mentor

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900 m is the entire distance. Note that the initial and final speeds are both zero over the 900 m.miscellaneous said:My method:

deltaS= (Vf^2-Vo^2)/2a

900= (Vf^2/.4) + (Vf^2)

900= 2.5Vf^2 + Vf^2

Vf= 16 m/s

Break the motion into two pieces, each with a different acceleration. Use that same kinematic formula, but for each piece separately. (The first piece goes from v = 0 to v = V1; the second goes from v = V1 to v = 0.)

- #3

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- #4

Doc Al

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Looks good. (I missed what you did, at first look.)miscellaneous said:

- #5

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Ohh, thanks for the confirmation. I'm really happy I know what I'm doing

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