Kinematics - What do these variables mean?

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kbn
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Homework Statement


first off, i don't completely understand what do the variables in the equation do:

x=Xo+volt+1/2at2

the exercise:
A body is thrown up vertically, it passes trough the height 9,8m twice. Time between those two passes is 4 seconds. Whats the starting velocity ?


Homework Equations





The Attempt at a Solution



i presume i have to do it with two equation systems? first off i'd have to calculate the starting velocity needed to get up and down in four seconds. Then i'd calculate the starting speed needed to get to the "seconds starting speed" at 9,8m and that would be the answer.

Bigthnx
 
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[tex]x=x_0 + v_0 t + \frac{1}{2} a t^2[/tex]
[tex]x[/tex] is displacement at time [tex]t[/tex]
[tex]x_0[/tex] is initial displacement
[tex]v_0[/tex] is initial velocity
[tex]a[/tex] is acceleration (constant)
[tex]t[/tex] is time

hint: [tex]a=9.8 m/s^2[/tex] in this case
 
great thanks for explaining the equation for me (y) everything seems much clearer now.

So i tried to solve it, but the answer isn't what its supposed to be =/ I mean, it different in the end of the book. Is the book wrong or am i ?

9,8=0 + Vo4 - 1/2*9,8*4[tex]^{2}[/tex]

4Vo = 1/2*9,8*4[tex]^{2}[/tex] + 9,8

Vo= (1/2*9,8*4[tex]^{2}[/tex] + 9,8) / 4 = 22,05 m/s

According to the textbook the answer should be 24 m/s

Thanks again