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 #?

2 Answers
Tom
Jan 6, 2016

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))#

Name
Jan 6, 2016

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#.

Explanation:

  • 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?

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