Problems with calculating the acceleration

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The discussion revolves around calculating the acceleration of a block moving 11 meters in 5 seconds, leading to confusion over three different calculated values. The key point is the distinction between average velocity and instantaneous velocity, with the average velocity being 11/5 m/s. It is clarified that since the block starts from rest, the initial velocity (V0x) is zero, which affects the choice of equations used for the calculation. The only applicable formula for this scenario is the one that incorporates distance, time, and acceleration without relying on instantaneous velocity. Ultimately, understanding these concepts resolves the discrepancies in the acceleration calculations.
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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).
 
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Assuming V0x = 0, then Vx ≠ 11/5 m/s.
 
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lewando said:
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?
 
11/5 m/s is the average velocity. Vx is the instantaneous velocity at the 5m point.
 
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lewando said:
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?
 
Yes! :smile:
 
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