We're asked to find the instantaneous velocity of an object, given its position equation.
In component equations, the position is
#x(t) = sin(2t-pi/2)#
#y(t) = (cost)/t#
The velocity can be found by differentiating the position equations:
#v_x(t) = d/(dt) [sin(2t-pi/2)] = 2sin(2t)#
#v_y(t) = d/(dt) [(cost)/t] = -(tsint + cost)/(t^2)#
To find the velocity components at #t = (3pi)/8#, we'll plug it in:
#v_x((3pi)/8) = 2sin(2((3pi)/8)) = 2sin((3pi)/4) = sqrt2# #"LT"^-1#
#v_y((3pi)/8) = -(((3pi)/8)sin((3pi)/8) + cos((3pi)/8))/(((3pi)/8)^2) = -1.06# #"LT"^-1#
The term "#"LT"^-1#" is there to remind us velocity is measured in units of distance divided by time, which in dimensional form is #"LT"^-1#. I wrote this because no units for either were given.
The magnitude of the instantaneous velocity is thus
#v((3pi)/8) = sqrt((sqrt2)^2 + (-1.06)^2) = color(red)(1.77# #color(red)("LT"^-1#
And the direction is
#phi = arctan((-1.06)/(sqrt2)) = color(blue)(-37.9^"o"#
with respect to the positive #x#-axis.