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?

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 .

- 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.