An object's two dimensional velocity is given by #v(t) = ( t^2 +2 , cospit - t )#. What is the object's rate and direction of acceleration at #t=2 #?

Question

An object's two dimensional velocity is given by #v(t) = ( t^2 +2 ,  cospit - t )#.  What is the object's rate and direction of acceleration at #t=2  #?

Charlotte Aaron · Accepted Answer

#a = 4.12# #"LT"^-2#

#phi = -14.0^"o"# We're asked to find the magnitude and direction of an object's acceleration at a certain time given its velocity equation. In component form, the velocity is #v_x(t) = t^2+2# #v_y(t) = cos(pit)-t# To find the acceleration as a function of time, we must differentiate the velocity equations: #a_x(t) = d/(dt) [t^2-2] = 2t# #a_y(t) = d/(dt) [cos(pit)-t] = -pisin(pit) - 1# At #t = 2# (ambiguous units here), we have #a_x(2) = 2(2) = 4# #"LT"^-2# #a_y(2) = -pisin(2pi) - 1 = -1# #"LT"^-2# (The term #"LT"^-2# is the dimensional form the units for acceleration, #"length"# #xx# #"time"^-2#. I used it here simply because no units were given.) The magnitude of the acceleration is #a = sqrt((a_x)^2 + (a_y)^2) = sqrt(4^2 + (-1)^2) = color(red)(4.12# #color(red)("LT"^-2# The direction is #phi = arctan((a_y)/(a_x)) = arctan((-1)/4) = color(blue)(-14.0^"o"# Always check the arctangent calculation to make sure your answer is not #180^"o"# off!

Benjamin Asiedu · Answer

undefined The object's rate of acceleration at $ t = 2 $ is $ (2, -\sin(2)) $ and its direction of acceleration is $ \theta = \tan^{-1}\left(\frac{-\sin(2)}{2}\right) $.