Problems with calculating the acceleration

[PhyIsOhSoHard]

Homework Statement

A block moves 11 meters in 5 seconds on a straight line. Calculate the acceleration.

Homework Equations

$v^2_x=v^2_{0x}+2a_x(x-x_0)$

$v_x=v_{0x}+a_xt$

$x=x_{0}+v_{0x}t+\frac{1}{2}a_{x}t^2$

The Attempt at a Solution

$a_x=\frac{(\frac{11m}{5s})^2}{2\cdot 11m}=0.22m/s^2$

$a_x=\frac{(\frac{11m}{5s})}{5s}=0.44m/s^2$

$a_x=\frac{11m}{1/2\cdot (5s)^2}=0.88m/s^2$

Why am I getting 3 different answers? In my textbook, it lists the equations as being practically the same (straight-line motion with constant acceleration).

 

[lewando]

Assuming V_0x = 0, then V_x ≠ 11/5 m/s.

 

[PhyIsOhSoHard]

Assuming V_0x = 0, then V_x ≠ 11/5 m/s.

yes, V_0x=0 because it starts at rest. But how do you know the velocity is not 11/5 m/s then?

 

[lewando]

11/5 m/s is the average velocity. V_x is the instantaneous velocity at the 5m point.

 

[PhyIsOhSoHard]

11/5 m/s is the average velocity. V_x is the instantaneous velocity at the 5m point.

Ohhhh.
And because of that, I cannot use any of the formulas which includes the V_x therefore the only correct equation for my current problem is the 3rd, correct?

 

[lewando]

Yes!