A speedboat starts from rest and accelerates at +2.01 m/s2 for 7.15 s. At the end of this time, the boat continues for an additional 6.25 s with an acceleration of +0.518 m/s2. Following this, the boat accelerates at -1.49 m/s2 for 8.55 s.
(a) What is the velocity of the boat at t = 21.95 s?
(b)(b) Find the total displacement of the boat.
To solve this problem, we can use the equations of motion, particularly the kinematic equation:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
For part (a), we need to find the velocity of the boat at t = 21.95 s. To do this, we can divide the problem into three segments: the first segment with an acceleration of +2.01 m/s^2, the second segment with an acceleration of +0.518 m/s^2, and the third segment with an acceleration of -1.49 m/s^2.
Segment 1:
Given:
Initial velocity (u1) = 0 m/s
Acceleration (a1) = +2.01 m/s^2
Time (t1) = 7.15 s
Using the kinematic equation,
v1 = u1 + a1*t1
v1 = 0 + (2.01)(7.15)
Segment 2:
Given:
Initial velocity (u2) = v1 (velocity at the end of segment 1)
Acceleration (a2) = +0.518 m/s^2
Time (t2) = 6.25 s
Using the kinematic equation,
v2 = u2 + a2*t2
v2 = v1 + (0.518)(6.25)
Segment 3:
Given:
Initial velocity (u3) = v2 (velocity at the end of segment 2)
Acceleration (a3) = -1.49 m/s^2
Time (t3) = 8.55 s
Using the kinematic equation,
v3 = u3 + a3*t3
v3 = v2 + (-1.49)(8.55)
Since we want to find the velocity at t = 21.95 s, we need to add the velocities at the end of each segment:
velocity at t = 21.95 s = v1 + v2 + v3
For part (b), we need to find the total displacement of the boat. To do this, we calculate the displacements for each segment using the formula:
s = ut + (1/2)at^2
Then, add the displacements from each segment:
total displacement = displacement from segment 1 + displacement from segment 2 + displacement from segment 3
Let's plug in the given values and calculate the solutions.