We're asked to find the magnitude (rate) and direction of the object's acceleration at #t = 1# #"s"#, given the object's velocity equation.
We need to find the object's acceleration as a function of time, doing so via differentiation of the velocity equation:
#veca = d/(dt) vecv#
#vecv = (t-2)hati + (2cos(pi/2t)-3t)hatj# (given, component form)
#v_x = t-2#
#a_x = 1# (derivative of #t-2#)
#v_y = 2cos(pi/2t)-3t#
#a_y = -pisin(pi/2t)-3# (derivative)
Now that we have the components of the acceleration as a function of #t#, let's plug in #1# for #t# to find the acceleration components at #t = 1# #"s"#:
#a_x = color(red)(1# #color(red)("m/s"^2#
#a_y = -pisin(pi/2(1))-3 = color(green)(-6.14# #color(green)("m/s"^2#
The magnitude of the acceleration at this time is
#a = sqrt((a_x)^2 + (a_y)^2) = sqrt((color(red)(1)color(white)(l)color(red)("m/s"^2))^2 + (color(green)(-6.14)color(white)(l)color(green)("m/s"^2))^2)#
#= color(blue)(6.22# #color(blue)("m/s"^2#
The direction of the acceleration is
#theta = arctan((a_y)/(a_x)) = arctan((color(green)(-6.14)color(white)(l)color(green)("m/s"^2))/(color(red)(1)color(white)(l)color(red)("m/s"^2))) = -1.41"rad" = color(purple)(-80.8^"o"#
