A particle moves in the x-y plane with velocity #v_x#=#8t-2# and #v_y#=2.If it passes through the points #x=14# and #y#=4 at #t#=2,what is the equation of the path?

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Answer 1 · 2018-03-18T20:28:31

#x = y^2-y+2#

Given #v_x(t) = 8t -2# and #v(t) = 2#, #x(2) = 14#, and #y(2) = 4#

Substitute #v_x(t) = (dx(t))/dt#

#(dx(t))/dt = 8t -2#

Integrating:

#x(t) = 4t^2-2t+x(0)" [1]"#

Substitute #x(2) = 14#

#14 = 4(2)^2-2(2)+x(0)#

#x(0) = 2#

Substitute into equation [1]:

#x(t) = 4t^2-2t+2" [1.1]"#

Substitute #v_y(t) = (dy(t))/dt#

#(dy(t))/dt = 2#

Integrate:

#y(t) = 2t+y(0)" [2]"#

Substitute #y(2) = 4#

#4 = 2(2) + y(0)#

#y(0) = 0#

Substitute into equation [2]:

#y(t) = 2t" [2.1]"#

Solve equation [2.1] for t:

#t = y/2#

Substitute into equation [1.1]:

#x = 4(y/2)^2-2(y/2)+2" [1.2]"#

Simplify:

#x = y^2-y+2" [1.3]"#