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

1 Answer
Jul 14, 2017

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

#phi = -37.9^"o"#

Explanation:

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.