Embarassing question- Displacement equation

[literacola]

Embarassing question-- Displacement equation

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

 

[AtticusFinch]

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

\Delta x = v_0t + \frac{1}{2}at^2

v = v_0 + at

 

[literacola]

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

 

[Borek]

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

 

[bolbol2054]

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

 

[Matterwave]

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!

 

[AtticusFinch]

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?

 

[AtticusFinch]

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.

a = \frac{dv}{dt}

a dt = dv

\int a dt = \int dv

at = v - v_0

at + v_0 = \frac{dx}{dt}

atdt + v_0dt = dx

\int atdt + \int v_0dt = \int dx

<br /> \Delta x = v_0t + \frac{1}{2}at^2 <br />

 

[zgozvrm]

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

 

[Jokerhelper]

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

 

[zgozvrm]

Yes, but so is the original question.