Embarassing question- Displacement equation

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Embarassing question-- Displacement equation

Homework Statement



Im just trying to understand why the equation for displacement along a line is [tex]x= 1/2(v_{0}+v)t[/tex]. If the other equation for displacement is [tex]v*t[/tex], where does the 1/2 come from?



Homework Equations


[tex]v*t[/tex]
[tex]x= 1/2(v_{0}+v)t[/tex]


The Attempt at a Solution


 
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Derive the formula you are asking about by using these other two kinematic equations

[tex]\Delta x = v_0t + \frac{1}{2}at^2[/tex]

[tex]v = v_0 + at[/tex]
 


Where does the 1/2 come from in the first equation you've given??
 


From integration needed to calculate displacement in uniformly accelerated linear motion.
 


you must take in consideration difference between Average velocity and Instantaneous velocity

x = vt v is average velocity

x= 1/2* (V0 + V )t now v is velocity at time t
 
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Perhaps it's easy to see that the two equations are valid for different situations. Let's just consider the situation of a constant velocity v.

You will notice that if I have a constant velocity v, then the v0 the second equation: .5(v0+v)t is simply v (since velocity is constant, the initial velocity is equal to the velocity always). Therefore the second equation just reduces to: .5(v+v)t=vt the first equation!
 


bolbol2054 said:
you must take in consideration difference between Average velocity and Instantaneous velocity

x = vt v is average velocity

x= 1/2* (V0 + V )t now v is velocity at time t

Yeah, the equation x = vt is not accurate for an object with an acceleration. Perhaps that is why you are confused?
 


literacola said:
Where does the 1/2 come from in the first equation you've given??

Borek said:
From integration needed to calculate displacement in uniformly accelerated linear motion.

[tex]a = \frac{dv}{dt}[/tex]

[tex]a dt = dv[/tex]

[tex]\int a dt = \int dv[/tex]

[tex]at = v - v_0[/tex]

[tex]at + v_0 = \frac{dx}{dt}[/tex]

[tex]atdt + v_0dt = dx[/tex]

[tex]\int atdt + \int v_0dt = \int dx[/tex]

[tex] \Delta x = v_0t + \frac{1}{2}at^2 [/tex]
 


To simplify, the average of 2 speeds is 1/2 their sum!
 


This is of course true only when dealing with a constant acceleration, or that at least isn't a function of time.
 


Yes, but so is the original question.