A speedboat starts from rest and accelerates at +1.69 m/s2 for 7.18 s. At the end of this time, the boat continues for an additional 4.55 s with an acceleration of +0.468 m/s2. Following this, the boat accelerates at -0.657 m/s2 for 8.38 s. (a) What is the velocity of the boat at t = 20.11 s? (b) Find the total displacement of the boat.
To solve this problem, we will use the equations of motion. Let's break down the problem step by step.
Step 1: Calculate the velocity at the end of each time interval.
Initial velocity, u = 0 m/s
(a) During the first time interval:
Acceleration, a1 = +1.69 m/s^2
Time, t1 = 7.18 s
Using the equation:
v = u + at
v1 = 0 + (1.69)(7.18)
v1 = 12.1522 m/s
(b) During the second time interval:
Acceleration, a2 = +0.468 m/s^2
Time, t2 = 4.55 s
Using the same equation:
v2 = v1 + (0.468)(4.55)
v2 = 14.7004 m/s
(c) During the third time interval:
Acceleration, a3 = -0.657 m/s^2
Time, t3 = 8.38 s
v3 = v2 + (-0.657)(8.38)
v3 = 8.95926 m/s
Step 2: Calculate the velocity at t = 20.11 s
t4 = 20.11 s
During the first time interval (t1), the velocity is constant at v1 = 12.1522 m/s.
Using the equation:
v4 = v1 + (a2)(t4 - t1)
v4 = 12.1522 + (0.468)(20.11 - 7.18)
v4 = 14.0902 m/s
So, the velocity of the boat at t = 20.11 s is 14.0902 m/s.
Step 3: Calculate the total displacement of the boat.
Displacement can be calculated using the equation:
s = ut + (1/2)at^2
(a) During the first time interval (0 to t1):
s1 = (1/2)(1.69)(7.18)^2
s1 = 56.0461 m
(b) During the second time interval (0 to t2):
s2 = (1/2)(0.468)(4.55)^2
s2 = 4.03358 m
(c) During the third time interval (0 to t3):
s3 = (1/2)(-0.657)(8.38)^2
s3 = -22.4408 m
(d) During the fourth time interval (t1 to t4):
s4 = v1(t4 - t1) + (1/2)(a2)(t4 - t1)^2
s4 = 12.1522(20.11 - 7.18) + (1/2)(0.468)(20.11 - 7.18)^2
s4 = 99.5769 m
Total displacement = s1 + s2 + s3 + s4
Displacement = 56.0461 + 4.03358 - 22.4408 + 99.5769
Displacement = 137.21678 m
So, the total displacement of the boat is 137.21678 m.
To solve this problem, we can use the kinematic equations of motion. These equations relate the displacement (d), initial velocity (Vi), final velocity (Vf), acceleration (a), and time (t).
(a) To find the velocity of the boat at t = 20.11 s, we need to calculate it in three different time intervals:
1. The first 7.18 s:
Using the equation Vf = Vi + at, where Vi is the initial velocity, a is the acceleration, and t is the time.
Vi = 0 m/s (the boat starts from rest)
a = +1.69 m/s^2 (positive because it's accelerating)
t = 7.18 s
Vf = 0 + (1.69 * 7.18) = 12.1542 m/s
2. The next 4.55 s:
Using the same equation:
Vi = 12.1542 m/s (the final velocity from the previous interval)
a = +0.468 m/s^2
t = 4.55 s
Vf = 12.1542 + (0.468 * 4.55) = 14.3172 m/s
3. The final 8.38 s:
Vi = 14.3172 m/s (the final velocity from the previous interval)
a = -0.657 m/s^2 (negative because it's decelerating)
t = 8.38 s
Vf = 14.3172 + (-0.657 * 8.38) = 8.0894 m/s
Therefore, at t = 20.11 s, the velocity of the boat is 8.0894 m/s.
(b) To find the total displacement of the boat, we need to calculate the displacement in each time interval and then sum them.
1. The displacement in the first 7.18 s:
Using the equation d = Vit + (1/2)at^2, where d is the displacement, Vi is the initial velocity, a is the acceleration, and t is the time.
Vi = 0 m/s
a = +1.69 m/s^2
t = 7.18 s
d = 0 * 7.18 + (1/2) * 1.69 * (7.18)^2 = 76.113 m
2. The displacement in the next 4.55 s:
Vi = 12.1542 m/s
a = +0.468 m/s^2
t = 4.55 s
d = 12.1542 * 4.55 + (1/2) * 0.468 * (4.55)^2 = 46.9811 m
3. The displacement in the final 8.38 s:
Vi = 14.3172 m/s
a = -0.657 m/s^2
t = 8.38 s
d = 14.3172 * 8.38 + (1/2) * (-0.657) * (8.38)^2 = 62.2997 m
Therefore, the total displacement of the boat is 76.113 m + 46.9811 m + 62.2997 m = 185.3938 m.
Therefore, at t = 20.11 s, the velocity of the boat is 8.0894 m/s and the total displacement is 185.3938 m.