# Further Kinematics understanding doubt

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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?

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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(Original post by

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?

**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?

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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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.

**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.

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Thankyou for replying. So for the second equation why does the equation not equal displacement vector while equation 1 equals the displacement vector?

**Stormragexox**)Thankyou for replying. So for the second equation why does the equation not equal displacement vector while equation 1 equals the displacement vector?

**both**equal displacement vectors. TBH it's easier to just rewrite them as:

... (*)

... (**)

then clearly the only difference is that in the second equation, we have zero acceleration, i.e. .

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

since acceleration is zero.

Then you also don't have in your first equation. This means the result is the displacement vector from the starting potision ... but including it into the equation means the result is displacement from the origin / some point of reference. If then there is no difference between these two cases.

Last edited by RDKGames; 2 years ago

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