A speedboat starts from rest and accelerates at +1.99 m/s2 for 7.00 s. At the end of this time, the boat continues for an additional 6.00 s with an acceleration of +0.515 m/s2. Following this, the boat accelerates at -1.44 m/s2 for 8.00 s. What is the velocity of the boat at t = 21.0 s?
V = a1*t1 + a2*t2 -a3*t3,
V = 1.99 * 7.0 + 0.515 * 6 - 1.44 * 8 = 13.93 + 3.09 - 11.52 = 5.5 m/s.
To find the velocity of the boat at t = 21.0 s, we need to break down the problem into smaller intervals and calculate the velocity at each interval using the given information.
Step 1: Calculate the velocity at the end of the first interval (from t = 0 to t = 7.00 s).
Using the equation v = u + at, where v is the final velocity, u is the initial velocity (which is zero since the boat starts from rest), a is the acceleration, and t is the time interval, we can calculate the velocity at the end of the first interval.
v1 = 0 + (1.99 m/s^2)(7.00 s) = 13.93 m/s
Step 2: Calculate the velocity at the end of the second interval (from t = 7.00 s to t = 13.00 s).
Using the same equation v = u + at, but now u is the velocity at the end of the first interval (v1) and a is the new acceleration (0.515 m/s^2), we can calculate the velocity at the end of the second interval.
v2 = 13.93 m/s + (0.515 m/s^2)(6.00 s) = 17.61 m/s
Step 3: Calculate the velocity at the end of the third interval (from t = 13.00 s to t = 21.00 s).
Again, using the same equation v = u + at, but now u is the velocity at the end of the second interval (v2) and a is the new acceleration (-1.44 m/s^2), we can calculate the velocity at the end of the third interval.
v3 = 17.61 m/s + (-1.44 m/s^2)(8.00 s) = 5.49 m/s
Therefore, the velocity of the boat at t = 21.0 s is 5.49 m/s.