Look of trajectory of a material point

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SUMMARY

The trajectory of a material point moving according to the equations x(t) = At², y(t) = Bt, and z(t) = 0 lies entirely in the x-y plane. The position vector is defined as r = At²*i + Bt*j. The velocity vector is derived as Velocity = 2At*i + B*j. The trajectory can be expressed in standard form as y = B*sqrt(x/A), confirming that it resembles a square root function rather than a logarithmic function.

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Chemist@
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Homework Statement


A body is moving, with the following data (these are all dependencies of the radius vector components respective to time): x(t)=At2, A>0; y(t)=Bt, B>0, z(t)=0. How does the trajectory of this movement look?

Homework Equations


r=x(t)i+y(t)j+z(t)k
Velocity=dr/dt

The Attempt at a Solution


r=At2*i+Bt*j (i,j and r are vectors)
Velocity=2At*i+B*j
Intuitively I think that the trajectory would look like a logarithm function, but is it correct and how to prove it?
 
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Since z(t)=0 for all t the trajectory must lie entirely in the x-y plane. Did you try to sketch the trajectory? Pick arbitrary values for A and B and plot a few points. Alternatively, find y as a function of x and comment on the type of function it represents.
 
You may know already a type of motion which satisfies similar conditions.
 
I thought that it would look similar to y=log(x). Correct?
 
Chemist@ said:
I thought that it would look similar to y=log(x). Correct?
Similar it may be, but you can find the actual equation by converting from parametric form to standard y(x) = ... form.
 
Is it y=B*sqrt(x/A)?
 
Chemist@ said:
Is it y=B*sqrt(x/A)?
Yup. Or if you combine the constants, ##y = c\sqrt{x}## .
 

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