Three start-ups Three cars, A, B, and C, start from rest

Problem 79E

Three start-ups Three cars, A, B, and C, start from rest and accelerate along a line according to the following velocity functions:

a. After t = 1 s, which car has traveled farthest?

b. After t = 5 s, which car has traveled farthest?

c. Find the position functions for the three cars assuming that all cars start at the origin

d. Which car ultimately gains the lead and remains in front?

Answer;

   Step 1 of  3 ;

          Given velocity functions are : , and .

1. In this problem we need to  find  after  t =1s,  which car has  traveled farthest.

          At  t = 1 seconds :  =  = 44

                                         =  = = 22

                                    =   = = 44

              Therefore , after t = 1s car A and car C  has traveled  farthest.

Step 2 of  3;

b) In this problem we need to  find  after  t =5s,  which car has  traveled farthest.

          At  t = 5 seconds :  =  = 73.333

                                         =  = = 61.1111

                                    =   = = 84.615

              Therefore , after t = 5s car C  has traveled  farthest.

Step 3 of 3 ;

               In this problem we need to find the  position functions  for  the three cars assuming  that  all cars  start at the origin.

Let us find the position function of  A :

                           (t)=  = 88dt

   By using partial fractions  we get :   = -

             Hence ,  (t) = 88dt = 88-)dt

                                                   = 88 (dt)

                                                   = 88( t - ln (1+t)) +C , since dx = ln |x| +c

   We know that , at initial point t = 0 .

  So ,  (0) = 0 = 88(  0 - ln (1+0)) +C

              0 = C , since ln(1) = 0

     Therefore , the position function of  A is :   (t) =  88(  t - ln (1+t)).

  Let us find the position function of  B :

                           (t) =  = 88dt

   By using partial fractions  we get :   =  + -

             Hence ,  (t) = 88dt  = 88+- )dt

                                                   = 88 (dt - dt)

                                                   = 88( t -  - 2 ln (1+t)) +C , since dx = ln |x| +c and  = n

   We know that , at initial point t = 0 .

  So , (0) = 0 = 88( 0 -  - 2 ln (1+0)) +C

              0 =  -88 +C  , since ln(1) = 0

            C = 88

     Therefore , the position function of  B is :   (t) =  88( t -  - 2 ln (1+t)) + 88

  Let us find the position function of  C :

                           (t) =  = 88dt

   By using partial fractions  we get :   =

             Hence ,  (t) = 88dt  = 88 -)dt

                                                   = 88 (dt)

                                                   = 88( t - ) +C , since dx = +c .   We know that , at initial point t = 0 .

  So , (0) = 0 = 88( 0 - ) +C

              0 =  0 +C  , since  = 0

            C = 0

     Therefore , the position function of  C is :   (t) =  88( t - )

d)  From  step-2  , it is clear that  car C  ultimately gains the lead and remains in front.