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Further Kinematics understanding doubt

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?
(edited 5 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.
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?
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:

r=r0+ut+12at2\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2 ... (*)

r=r0+ut\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. a=0\mathbf{a} = \mathbf{0}.

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

r=r0+vt\mathbf{r} = \mathbf{r}_0 + \mathbf{v} t

since acceleration is zero.

Then you also don't have r0\mathbf{r}_0 in your first equation. This means the result r\mathbf{r} is the displacement vector from the starting potision r0\mathbf{r}_0... but including it into the equation means the result is displacement from the origin / some point of reference. If r0=0\mathbf{r}_0 = \mathbf{0} then there is no difference between these two cases.
(edited 5 years ago)

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