To solve this problem, we can use the equations of motion. Let's break down the problem into different intervals and calculate the velocity and displacement at each interval.
Interval 1: From t=0 to t=8.25 s with an acceleration of +1.98 m/s^2
Using the equation of motion: v = u + at
We know the initial velocity u = 0 (starting from rest), acceleration a = +1.98 m/s^2, and time t = 8.25 s.
Substituting the values, we get:
v1 = 0 + (1.98)(8.25)
v1 = 16.335 m/s
Interval 2: From t=8.25 s to t=13.83 s with an acceleration of +0.348 m/s^2
Using the same equation of motion:
v2 = v1 + a*t2
Where v1 is the velocity at the end of the first interval, a is the acceleration, and t2 is the time for the second interval.
Substituting the values:
v2 = 16.335 + (0.348)(5.58)
v2 = 18.795 m/s
Interval 3: From t=13.83 s to t=21.05 s with an acceleration of -1.42 m/s^2
Using the same equation of motion:
v3 = v2 + a*t3
Where v2 is the velocity at the end of the second interval, a is the acceleration, and t3 is the time for the third interval.
Substituting the values:
v3 = 18.795 + (-1.42)(7.22)
v3 = 7.3414 m/s
(a) Velocity at t=21.05 s:
The velocity at this time is given by v3, which we just calculated:
v = v3
v = 7.3414 m/s
(b) Total displacement:
The total displacement is the sum of the individual displacements during each interval. We can calculate it using the equation:
s = ut + (1/2)at^2
For each interval, we need to calculate the displacement separately, and then add them up.
Displacement during interval 1:
s1 = (0)(8.25) + (1/2)(1.98)(8.25)^2
s1 = 68.07675 m
Displacement during interval 2:
s2 = (16.335)(5.58) + (1/2)(0.348)(5.58)^2
s2 = 136.50384 m
Displacement during interval 3:
s3 = (18.795)(7.22) + (1/2)(-1.42)(7.22)^2
s3 = 76.837276 m
Total displacement:
s_total = s1 + s2 + s3
s_total = 68.07675 + 136.50384 + 76.837276
s_total = 281.417066 m
Therefore, the velocity at t=21.05 s is 7.3414 m/s and the total displacement of the boat is 281.417066 m.