How Do You Calculate the Velocity Vector When a Particle's Position is x=25?

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SUMMARY

The velocity vector of a particle moving in the xy-plane, defined by the position functions x(t)=5t+3sin(t) and y(t)=(8-t)(1-cos(t)), is calculated when the horizontal position x equals 25. The solution requires solving the equation 5t + 2sin(t) = 25 numerically, as it cannot be solved algebraically. The final velocity vector at this position is determined to be (7.008, -2.228).

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A particle moves in the xy-plane so that the position of the particle is given by
x(t)=5t+3sint , y(t)=(8-t)(1-cost)

Find the velocity vector at the time when the particle's horizontal position is x=25
answer is (7.008, -2.228)

*the x=25, i think the trick is somehow get the t with that..
 
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Yes, it is. You have to solve 5t+ 2sin(t)= 25. There is no "algebraic" way to solve an equation like that. You will have to do it numerically. If you have a graphing calculator, you could see where the graphs of y= 5t+ 2sin(t) crosses the graph of y= 25.

If not, you might use a "bisection" method. For example, It is obvious that X(4)= 20+ 3sin(4)< 25. It is not so obvious but easy to show with a calculator that X(7)= 28+ 3 sin(7)> 25 (but not 5 or 6). The function must be equal to 25 some place between them. Just because it is simplest, try half way between them, 5.5. X(5.5)= 25.3> 25. Now we know the function is equal to 25 some place between 4 and 5.5. Again, try half way: (4+ 5.5)/2= 4.75. X(4.75)= 20.817< 25 so now we know the solution is somewhere between 4.75 and 5.5. Half way between them is (4.75+ 5.5)/2= 5.125.

Continue unfil you feel you have sufficient accuracy.
 

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