What is the instantaneous velocity of an object moving in accordance to # f(t)= (sin(2t-pi/2),cost/t ) # at # t=(3pi)/8 #?

Question

What is the instantaneous velocity of an object moving in accordance to #              f(t)=   (sin(2t-pi/2),cost/t )   # at #  t=(3pi)/8      #?

Isabella Ackerman · Accepted Answer

#v((3pi)/8) = 1.77# #"LT"^-1#

#phi = -37.9^"o"# 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.

Isaiah Allen · Answer

undefined The instantaneous velocity of the object at $ t = \frac{3\pi}{8} $ can be found by taking the derivative of the function $ f(t) $ with respect to $ t $, then evaluating it at $ t = \frac{3\pi}{8} $. Given the function $ f(t) = (\sin(2t-\frac{\pi}{2}), \frac{\cos(t)}{t}) $, the instantaneous velocity can be calculated as follows:

$$ f'(t) = \left(2\cos(2t-\frac{\pi}{2}), -\frac{\sin(t)}{t} - \frac{\cos(t)}{t^2}\right) $$

Now, substitute $ t = \frac{3\pi}{8} $ into $ f'(t) $ to find the instantaneous velocity vector at $ t = \frac{3\pi}{8} $.