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A particle's position is r(vector) =(ct^2−2dt^3)i^+(2ct^2−dt^3)j^,. where c and d are positive constants. Find expressions for times t > 0 when the particle is moving in the x-direction. Express your answer in terms of the variables c and d.

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It has to be understood that when the particle is moving in the direction of x, then y must be zero.
Then
r = (ct^2−2dt^3) i^ 
Now, differentiating "r" twice, we get
v = 2ct - 6dt^2 
a = 2c - 12dt 
12dt = 2c - a 
t = [2c-a]/12d

For the second case
 r = (2ct^2−dt^3) j^ 
Now, differentiating 'r' twice, we get
v = 4ct - 3dt^2 
a = 4c - 6dt 
6dt = 4c - a 
t = [4c-a]/6d

I hope that the answer has come to your help. 

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