What is the instantaneous velocity of an object moving in accordance to # f(t)= (sqrt(t+2),t+4) # at # t=1 #?

Question

What is the instantaneous velocity of an object moving in accordance to #              f(t)=   (sqrt(t+2),t+4)    # at #  t=1    #?

Samantha Adkison · Accepted Answer

#sqrt 2# Take the derivative of #sqrt (t+2)#. You should get: #1/(2sqrt(t+2))# Plug in 1 to get #1/2sqrt(2)#. Now take the derivative of #t+4#. You should get 1 so plugging in 1 will get you 1. Now that you know that it has a velocity of #1/2sqrt(2)#. in the x direction and 1 in the y direction, you need to find the resulting velocity from that. Use Pythagorean theorem: #a^2+b^2=c^2# #(1/(2sqrt2))^2+1^2=c^2# #1/(4*2)+1=c^2# #c=sqrt(9/8)#

Samantha Adkison · Answer

Instantaneous velocity is given by the vector:

# vec(v) = << sqrt(3)/6,1 >> #, or, #sqrt(3)/6 hat(i)+hat(j)#, or, #( (sqrt(3)/6),(1) )#

We have: # f(t) = ( x(t), y(t) )# where #x(t)=sqrt(t+1)#, #y(t)=t+4# Then; # dx/dt = 1/2(t+2)^(-1/2) = 1/(2sqrt(t+2))# # dy/dt = 1 # So, when #t=1 #: # dx/dt = 1/(2sqrt(1+2)) = 1/(2sqrt(3)) = sqrt(3)/6# # dy/dt = 1 # And so the instantaneous velocity is given by the vector: # vec(v) = << sqrt(3)/6,1 >> #, or, #sqrt(3)/6 hat(i)+hat(j)#, or, #( (sqrt(3)/6),(1) )# If we want the instantaneous speed , it is given by: # v = || vec(v) || # # \ \ \= sqrt( 3/36 + 1 ) # # \ \ \= sqrt( 13/12) # # \ \ \~~ 1.04083 #

Emilia Aguilar · Answer

undefined The instantaneous velocity of an object moving according to the function $ f(t) = (\sqrt{t+2}, t+4) $ at $ t = 1 $ can be found by taking the derivative of the position function with respect to time and then evaluating it at $ t = 1 $.

The derivative of $ f(t) = (\sqrt{t+2}, t+4) $ with respect to $ t $ is $ f'(t) = \left(\frac{1}{2\sqrt{t+2}}, 1\right) $.

Substituting $ t = 1 $ into the derivative, we get $ f'(1) = \left(\frac{1}{2\sqrt{1+2}}, 1\right) = \left(\frac{1}{2\sqrt{3}}, 1\right) $.

Therefore, the instantaneous velocity of the object at $ t = 1 $ is $ \left(\frac{1}{2\sqrt{3}}, 1\right) $.