What is the instantaneous velocity of an object with position at time t equal to # f(t)= (tsqrt(t+2),t^2-2t) # at # t=1 #?

Question

What is the instantaneous velocity of an object with position at time t equal to #              f(t)=   (tsqrt(t+2),t^2-2t)  # at #  t=1  #?

Emma Aldridge · Accepted Answer

We have the parametric curve which define the movement of your material point #M# We need to derive once to have the parametric curve which define the velocity of your material point #M# let's do that : #x(t) = tsqrt(t+2)#
#y(t) = t^2-2t# #vec(OM)(t) = (x(t),y(t))# #dx/dt= dot(x) = v_x(t) = sqrt(t+2)+t/(2sqrt(t+2)) = (3t+4)/(2sqrt(t+2))# #dy/dt = dot(y) = v_y(t) = 2t-2# #(dvec(OM))/dt = vec(v)(t) = (v_x(t),v_y(t)) = ((3t+4)/(2sqrt(t+2)),2t-2) # #vec(v)(1) = (7/(2sqrt(3)),0)# Your material point #M# doesn't have a movement relative to the y-axis, so the norm is #v(1) = 7/(2sqrt(3))#

Ezekiel Alexander · Answer

If you calculate the function of distance in time #f_d(t)#from the given function of position in time #f(t)#, you can derive that the instantaneous velocity of the object when #t=1# is #7/4#.

The instantaneous velocity of an object is the derivative of distance as a function of time. The distance traveled by the object can be taken as its the distance between (0,0) and the current position of the object. If distance as a function of time is #y=f_d(t)#, velocity is denoted #v#, the function of position in time is #f(t)# then: #v=(dt)/(dy)# #f_d(t)= sqrt((t^2-2t)^2+(tsqrt(t+2))^2)# #f_d(t)=sqrt((t^4-4t^3+4t^2)+(t^3+2t^2))# #f_d(t)=sqrt(t^4-3t^3+6t^2)=(t^4-3t^3+6t^2)^(1/2)# #v=(dt)/(dy)=d((t^4-3t^3+6t^2)^(1/2))/(dy)# #v=1/2[(t^4-3t^3+6t^2)^(-1/2)][4t^3-9t^2+12t]# #v=(4t^3-9t^2+12t)/(2sqrt(t^4-3t^3+6t^2))# #v=(4t^2-9t+12)/(2sqrt(t^2-3t+6))# When #t=1# #v=(4-9+12)/(2sqrt(1-3+6))# #v=7/4# I think?

Adam Adkins · Answer

undefined To find the instantaneous velocity of an object at $ t = 1 $, we need to find the derivative of the position function with respect to time, and then evaluate it at $ t = 1 $.

Given the position function $ f(t) = (t\sqrt{t + 2}, t^2 - 2t) $, we need to find $ f'(t) $ and then evaluate it at $ t = 1 $.

To find $ f'(t) $, we differentiate each component of the position function with respect to time $ t $.

The derivative of $ t\sqrt{t + 2} $ with respect to $ t $ is:
$$ \frac{d}{dt} (t\sqrt{t + 2}) = \sqrt{t + 2} + \frac{t}{2\sqrt{t + 2}} $$

The derivative of $ t^2 - 2t $ with respect to $ t $ is:
$$ \frac{d}{dt} (t^2 - 2t) = 2t - 2 $$

Now, we evaluate these derivatives at $ t = 1 $:

For the first component:
$$ \sqrt{1 + 2} + \frac{1}{2\sqrt{1 + 2}} = \sqrt{3} + \frac{1}{2\sqrt{3}} $$

For the second component:
$$ 2(1) - 2 = 0 $$

So, the instantaneous velocity of the object at $ t = 1 $ is $ (\sqrt{3} + \frac{1}{2\sqrt{3}}, 0) $.