Further Kinematics understanding doubt

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Stormragexox
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Hey guys I was just having a read through my text book and noticed that when using r0 + vt=r postion vector and when using suvat ut+1/2at^2= r( displacement at time t)

what is the difference between the two rs is the r for the second equation displacement?
Last edited by Stormragexox; 2 years ago
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mqb2766
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(Original post by Stormragexox)
Hey guys I was just having a read through my text book and noticed that when using r0 + vt=r postion vector and when using suvat ut+1/2at^2= r( displacement at time t)

what is the difference between the two rs is the r for the second equation displacement?
r = ut+1/2at^2
assumes constant acceleration "a", initial velocity "u" and the initial displacement (at time t = 0) is r = 0. You could add an r0 into this expression if the initial displacement is non-zero.

r = r0 + vt
assumes (constant) zero acceleration, the velocity is constant "v" and the initial displacement is "r0". Assuming a non-zero acceleration would give the first (or rearranged) equation.
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Stormragexox
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(Original post by mqb2766)
r = ut+1/2at^2
assumes constant acceleration "a", initial velocity "u" and the initial displacement (at time t = 0) is r = 0. You could add an r0 into this expression if the initial displacement is non-zero.

r = r0 + vt
assumes (constant) zero acceleration, the velocity is constant "v" and the initial displacement is "r0". Assuming a non-zero acceleration would give the first (or rearranged) equation.
Thankyou for replying. So for the second equation why does the equation not equal displacement vector while equation 1 equals the displacement vector?
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RDKGames
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(Original post by Stormragexox)
Thankyou for replying. So for the second equation why does the equation not equal displacement vector while equation 1 equals the displacement vector?
They both equal displacement vectors. TBH it's easier to just rewrite them as:

\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2 ... (*)

\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t ... (**)

then clearly the only difference is that in the second equation, we have zero acceleration, i.e. \mathbf{a} = \mathbf{0}.

Your use of \mathbf{u} and \mathbf{v} as different things doesn't apply since we don't have context... all we need to know is that \mathbf{u} is the velocity. If you want to introduce context, then we just use the equation \mathbf{v} = \mathbf{u} + \mathbf{a} t and sub it into (**) to get:

\mathbf{r} = \mathbf{r}_0 + \mathbf{v} t

since acceleration is zero.

Then you also don't have \mathbf{r}_0 in your first equation. This means the result \mathbf{r} is the displacement vector from the starting potision \mathbf{r}_0... but including it into the equation means the result is displacement from the origin / some point of reference. If \mathbf{r}_0 = \mathbf{0} then there is no difference between these two cases.
Last edited by RDKGames; 2 years ago
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