# Problems with calculating the acceleration

1. Sep 15, 2013

### PhyIsOhSoHard

1. The problem statement, all variables and given/known data
A block moves 11 meters in 5 seconds on a straight line. Calculate the acceleration.

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

$v_x=v_{0x}+a_xt$

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

3. 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).

2. Sep 15, 2013

### lewando

Assuming V0x = 0, then Vx ≠ 11/5 m/s.

3. Sep 15, 2013

### PhyIsOhSoHard

yes, V0x=0 because it starts at rest. But how do you know the velocity is not 11/5 m/s then?

4. Sep 15, 2013

### lewando

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

5. Sep 15, 2013

### PhyIsOhSoHard

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

6. Sep 15, 2013

Yes!