world class speed skaters can skate a 3,000 meter course in about 4 minutes. what is their average speed for this course?

12.5 m/s.

v = d/t

(60 s/1 min)*(4 min) = 240 s

v = 3000 m/240 s  

v = 12.5 m/s  

(a) 52.724 m/s

(b) Total displacement, d = 551.25 m

Solution:

As per the question:

Initial acceleration of the speed boat, a = - 2.01

Time duration, t = 7.00 s

Additional time, t' =6.00 s

Acceleration for additional time, a' = 0.518

The followed up acceleration, a'' = 1.49

Time duration, t'' = 8.00 s

(a) Now, to calculate the velocity of the boat at timer, t = 21.0 s, we have:

After the initial 7.00 s, the velocity of the boat, from eqn-1 of motion:

v = u + at

v = 0 - 2.1(7.00) = - 14.7 m/s

After t + t' = 13 s:

v' = v + at

v' =  14.7 + 0.518(13) =  21.434 m/s

Now, velocity of the boat after t = 21 s:

v'' = v' + a''t

v'' = 21.434 + 1.49(21) = 52.724 m/s

(b) Now, the total displacement, d:

For the first case:

d = ut +

For the second case:

d = v't' = 21.434(6) = 128.6 m

For the third case:

d = ut +

For the fourth case:

d = v''t'' = 52.724(8) = 421.79 m

For the last case:

d = ut +

Total displacement, d = -51.45 + 128.6 + 4.65 + 421.79 + 47.68 = 551.25 m