A particle of mass m moves in the xy plane; its acceleration in m/s^2 is given b

Question

A particle of mass m moves in the xy plane; its acceleration in m/s^2 is given by the expression, vecotr a = -6i for t greater than or equal to 0. At t=0 seconds, the object is at the point (Xi,Y,) = (36,0). At te4 seconds the object is at point (36,40). A) At which time (if any) in seconds is the particle at its maximum possible value of x? If there is no such maximum, state why. B) At what time (if any) in seconds is the particle moving parallel to the x-axis? If never, explain why. C) Find the time t greater than 0 in seconds at which the particle crosses the y-axis I know I need to integrate the acc. vector to get V(t) and r(t) but I don't know how to find the constants or even know where to begin

Explanation / Answer

A) a = -6t v = -3t^2i + (cxi + cy j) and r = t^3i + (cxi + cy j) t+ (dxi + dyj) so r = (t^3 + cxt + dx)i + (cyt + dy) j at t = 0 r = 36i + 0j dx = 36 and dy = 0 so r = (t^3 + cxt + 36)i + (cyt)j at t = 4 r = 36i + 40j 64 + 4cx + 36 = 36 and 4cy = 40 cx = -16 and cy = 10 so r = (t^3 - 16t + 36)i + (10t)j or r = (t^3 - 16t + 36, 10t ) now i have given u xpression of r an values of constants

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