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

##### 2 Answers

#### Explanation:

Take the derivative of

Plug in 1 to get

Now take the derivative of

Now that you know that it has a velocity of

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

# 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 #