Two balls have a race over a distance of 20.0 meters. The red ball travels at a constant velocity while the blue ball starts at rest and has a constant acceleration. For this problem set the blue ball's acceleration to be 3.30 m/s2. Note that the simulation may stop slightly after one of the balls reaches the finish line rather than at the instant it reaches the line.
a. How far is the second ball from the finish line at the instant the first ball reaches the line?
b. Assume the motion continues for both balls until the second ball reaches the finish line. What is the distance between the two balls at that time?

a) 2.592 m

b) 6.4 m

Explanation:

Given:

- The acceleration of blue ball a = 3.30 m/s^2

- The total track length s = 20.0 meters

- The initial velocity of red ball v_i = 5 m/s

Find:

a. How far is the second ball from the finish line at the instant the first ball reaches the line?

b. Assume the motion continues for both balls until the second ball reaches the finish line. What is the distance between the two balls at that time?

Solution:

- The amount of time taken by each ball to travel a distance of track s = 20 m is:

             Blue ball:         s = 0.5*a*t^2

                                     20 = 0.5*3.3*t^2

                                      t = sqrt ( 400 / 33 ) = 3.482 s

             Red ball:          s = v_o*t

                                     20 = 5*t

                                      t =  20 / 5 = 4 s

- Hence, blue ball reaches the distance of s = 20 m first @ t = 3.482s. The distance traveled by the Red ball in that time is:

             Red ball @t=3.482

                                       s = 5*(3.482) = 17.4078 m

- So at the end the distance between both balls is : 20 - 17.4078 = 2.592 m

- If the simulation continues, till Red ball reaches the finish line of track at s = 20 m in time t = 4s. The amount of distance traveled by the blue ball is:

                                      s = 0.5*3.3*(4)^2

                                      s = 26.4 m

- So at the end the distance between both balls is : 26.4 - 20 = 6.4 m

the diagram which shows forces as either balanced = same speed or unbalanced = accelerarting

a. are