Problems with calculating the acceleration

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Homework Help Overview

The discussion revolves around calculating the acceleration of a block moving a certain distance over a specified time. The original poster presents multiple attempts at using kinematic equations to find the acceleration but arrives at different results, leading to confusion regarding the correct approach.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply various kinematic equations but questions the validity of their results. Participants discuss the distinction between average and instantaneous velocity, which leads to further questioning about the assumptions made in the calculations.

Discussion Status

Participants are actively engaging in clarifying concepts related to average versus instantaneous velocity. Some guidance has been provided regarding the use of specific equations based on these distinctions, although there is no explicit consensus on the correct approach yet.

Contextual Notes

There is an assumption that the block starts from rest, which influences the interpretation of the velocities involved. The original poster is also navigating the constraints of using different kinematic equations in the context of the problem.

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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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