Problems with calculating the acceleration

  • #1

Homework Statement


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


Homework Equations


[itex]v^2_x=v^2_{0x}+2a_x(x-x_0)[/itex]

[itex]v_x=v_{0x}+a_xt[/itex]

[itex]x=x_{0}+v_{0x}t+\frac{1}{2}a_{x}t^2[/itex]

The Attempt at a Solution


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

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

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

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).
 

Answers and Replies

  • #2
lewando
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Assuming V0x = 0, then Vx ≠ 11/5 m/s.
 
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  • #3
Assuming V0x = 0, then Vx ≠ 11/5 m/s.
yes, V0x=0 because it starts at rest. But how do you know the velocity is not 11/5 m/s then?
 
  • #4
lewando
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11/5 m/s is the average velocity. Vx is the instantaneous velocity at the 5m point.
 
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  • #5
11/5 m/s is the average velocity. Vx is the instantaneous velocity at the 5m point.
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
lewando
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Yes! :smile:
 
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