two baseball players one on 1st base Player 1 and one on 2nd

“two baseball players, one on 1st base (Player 1) and one on 2nd base (Player 2). Player 1 runs at 18 ft/sec. Player 2 runs at 26 ft/sec. The distance between bases is 90 ft. The runners leave their bases simultaneously. Assume that they accelerate immediately to their running speeds. Runner 2 stops at 3rd base the moment he reaches it. Runner 1 stops at 2nd base the moment he reaches it.”

“a.Define a function B that gives the distance between Player 1 and 2nd base as a function of the number of seconds since they began running. You will need to define your function in two parts (ctrl-shift-A), one part for Player 1’s distance from 2nd base while he is running and another part for the distance after he reaches 2nd base.

b.Define a function A that gives the distance between Player 2 and 2nd base as a function of the number of seconds since they began running. You will need to define your function in two parts (ctrl-shift-A), one part for Player 2’s distance from 2nd base while he is running and another part for the distance after he reaches 3rd base.”

“c.Define a function D that gives the distance between runners as a function of the number of seconds since they began running.

d.Graph your function D in GC. (You will need to rescale your axes to see the graph.) Use GC’s tracing feature (click and drag on the graph) to estimate the maximum distance and the minimum distance between runners, and the number of seconds since starting at which these distances happen. If you define your functions correctly, the minimum distance will be 73.9973 feet and the maximum distance will be 94.16403 feet.

e.This is a challenge problem. Assume that each player accelerates to top speed at the rate of 30 (ft/sec)/sec. Rework all of a-d with this assumption.”

Solution

a. Player 1 is running towards 2nd Base as velocity = 18 ft/sec, DIstance between the bases is 90ft. Hence, Player 1 is going to complete the run in 90/18 = 5 sec (assuming instant reaching of the top speed)

B is the function that gives distance between the Player 1 and 2nd Base, then

at t= 0, B=90, at t = 5, B= 0

B= 90-18t, (0<=t<=5 sec)

b. Similary for 2nd Player, with velocity = 26 ft/sec, he is going to reach 3rd Base in 90/24 = 3.75 sec at t= 0, A = 0

A = 24t (0<=t<=3.75 sec)

c. Since both functions A & B are calculated from 2nd Base,

Distance between the two Players shall be D=A+B

D = 24t +90 - 18t = 90 + 6t (0 <=t<=3.75 sec)

After 3.75 sec, Player 2 has reached 3rd Base = 90m and has stopped hence

D = 90+90-18t = 180-18t (3.75<t<=5 sec)

d. Needs Graph

e. if the acceleration of 30 ft/sec² is to be used then

For Player 1, v = u + at

18 = 0 + 30t, at t = 0.6sec Player 2 reaches his maximum speed, By this times he has completed distance = 1/2at² = 15(0.6)(0.6) =  5.4ft. Remaining distance = 84.6ft to be completed at speed 18ft/sec requires t = 4.7 sec. Total time required to reach 2nd Base = 5.3sec

Player 2 reaches his maximum speed at t = 24/30 = 0.8 sec. By this times he has completed distance = 1/2at² = 15(0.8)(0.8) = 9.6ft. Remaining distance = 80.4ft to be completed at speed 24ft/sec requires t = 3.35 sec. Total time required to reach 3rd Base = 4.15 sec

B = 90 - 18t - 15t² ( 0<=t<=0.6sec)

B = 90 - 5.4 - 18(t-0.6) = 95.4 -18t (0.6<t<=5.3 sec)

A= 24t + 15t² (0<=t<=0.8sec)

A = 9.6 + 24(t-0.8) = 24t - 9.6 (0.8<t<=4.15 sec)

D = A + B (divided over times0, 0.6, 0.8, 4.15 & 5.3 sec)

i. for 0<-t<=0.6sec, D = 24t+15t² + 90 - 18t - 15t² = 90+6t

ii for 0.6<t<=0.8sec, D = 24t+15t² + 95.4 - 18t = 95.4 + 6t + 15t²

iii for 0.8<t,+4.15 sec, D = 24t - 9.6 + 95.4 - 18t = 85.8 + 6t

iv for 4.15<t<=5.3sec, D = 90 + 95.4 - 18t = 185.4 - 18t