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

2 Answers
Astralboy
Feb 18, 2017

#sqrt 2#

Explanation:

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)#

Steve M
Feb 18, 2017

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) )#

Explanation:

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 #

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