lets just look at x direction..
d(t)=vix*t+1/2 a t^2
dx(t)=4.8t-1/2 *2.5*t^2
now to find the max
d'/dt=0=4.8-2.5 t or max occurs at
time=4.8/2.5
now solve for position at this time:
d(t)= (4.8*t -1/2 *2.5*t^2) i^ + 4.3t^2 j^
and finally, velocity (derivitative of d(t)
v(t)=(4.8 -2.5t)i^ +8.3t j^
A particle starts from the origin at t = 0 with an initial velocity of 4.8m/s along the positive x axis.If the acceleration is (-2.5 i^ + 4.3 j^)m/s2, determine (a)the velocity and (b)position of the particle at the moment it reaches its maximum x coordinate
Express your answer using two significant figures. Enter your answers numerically separated by a comma.
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