A speedboat starts from rest and accelerates at +2.01 m/s2 for 7.05 s. At the end of this time, the boat continues for an additional 6.15 s with an acceleration of +0.518 m/s2. Following this, the boat accelerates at -1.49 m/s2 for 8.50 s.
(a) What is the velocity of the boat at t = 21.7 s?
m/s
(b) Find the total displacement of the boat.
find the initial and final velocity for each time period.
vf=vi+at
then each vf becomes vi for the next period.
For total displacment find d for each period, and add them
d=vi*t+1/2 a t^2
(a) Well, let's break it down step by step. First, let's find the velocity at the end of each time interval.
For the first interval (t = 0 to 7.05 s), the boat starts from rest and accelerates at +2.01 m/s² for 7.05 s. To find the velocity at the end of this interval, we can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time. Plugging in the values, we have:
v₁ = 0 + (2.01 m/s²)(7.05 s)
v₁ = 14.1505 m/s (approx)
For the second interval (t = 7.05 s to 13.20 s), the boat continues with an acceleration of +0.518 m/s². Since we already know the initial velocity at the start of this interval, we can use the same equation:
v = u + at
v₂ = v₁ + (0.518 m/s²)(6.15 s)
v₂ ≈ 17.3157 m/s (approx)
For the third interval (t = 13.20 s to 21.70 s), the boat accelerates at -1.49 m/s². Once again, using the equation:
v = u + at
v₃ = v₂ + (-1.49 m/s²)(8.50 s)
v₃ ≈ 4.2772 m/s (approx)
So, the velocity of the boat at t = 21.7 s is approximately 4.2772 m/s.
(b) To find the total displacement of the boat, we need to find the sum of the displacements during each interval.
The displacement, s, in each interval can be found using the equation:
s = ut + (1/2)at²
For the first interval (t = 0 to 7.05 s):
s₁ = (1/2)(2.01 m/s²)(7.05 s)²
s₁ ≈ 50.8003 m (approx)
For the second interval (t = 7.05 s to 13.20 s):
s₂ = v₁(6.15 s) + (1/2)(0.518 m/s²)(6.15 s)²
s₂ ≈ 106.3596 m (approx)
For the third interval (t = 13.20 s to 21.70 s):
s₃ = v₂(8.50 s) + (1/2)(-1.49 m/s²)(8.50 s)²
s₃ ≈ -47.0425 m (approx)
To find the total displacement, we add up the individual displacements:
total displacement = s₁ + s₂ + s₃
total displacement ≈ 110.1174 m (approx)
So, the total displacement of the boat is approximately 110.1174 m.
To find the velocity of the boat at t = 21.7 s, we need to calculate the total velocity after each acceleration phase.
Step 1: Calculate the velocity at the end of the first phase (7.05 s) using the formula:
v = u + at
Where:
v = final velocity
u = initial velocity (which is 0 since the boat starts from rest)
a = acceleration (+2.01 m/s^2)
t = time (7.05 s)
v1 = 0 + (2.01)(7.05)
v1 ≈ 14.1355 m/s
Step 2: Calculate the velocity at the end of the second phase (6.15 s) using the same formula:
v2 = v1 + a2t2
Where:
v2 = final velocity at the end of the second phase
v1 = velocity at the end of the first phase (14.1355 m/s)
a2 = acceleration (+0.518 m/s^2)
t2 = time (6.15 s)
v2 = 14.1355 + (0.518)(6.15)
v2 ≈ 17.9999 m/s
Step 3: Calculate the velocity at the end of the third phase (8.50 s) using the same formula:
v3 = v2 + a3t3
Where:
v3 = final velocity at the end of the third phase
v2 = velocity at the end of the second phase (17.9999 m/s)
a3 = acceleration (-1.49 m/s^2)
t3 = time (8.50 s)
v3 = 17.9999 + (-1.49)(8.5)
v3 ≈ 4.015 m/s
Therefore, at t = 21.7 s, the velocity of the boat is approximately 4.015 m/s.
To find the total displacement of the boat, we need to calculate the displacement during each phase and then add them together.
Step 1: Calculate the displacement during the first phase using the formula:
s1 = ut + 0.5at^2
Where:
s1 = displacement during the first phase
u = initial velocity (which is 0 since the boat starts from rest)
a = acceleration (+2.01 m/s^2)
t = time (7.05 s)
s1 = (0)(7.05) + 0.5(2.01)(7.05)^2
s1 ≈ 102.365 m
Step 2: Calculate the displacement during the second phase using the same formula:
s2 = v1t2 + 0.5a2t2^2
Where:
s2 = displacement during the second phase
v1 = velocity at the end of the first phase (14.1355 m/s)
a2 = acceleration (+0.518 m/s^2)
t2 = time (6.15 s)
s2 = (14.1355)(6.15) + 0.5(0.518)(6.15)^2
s2 ≈ 54.214 m
Step 3: Calculate the displacement during the third phase using the same formula:
s3 = v2t3 + 0.5a3t3^2
Where:
s3 = displacement during the third phase
v2 = velocity at the end of the second phase (17.9999 m/s)
a3 = acceleration (-1.49 m/s^2)
t3 = time (8.50 s)
s3 = (17.9999)(8.50) + 0.5(-1.49)(8.50)^2
s3 ≈ 135.299 m
Therefore, the total displacement of the boat is approximately 102.365 m + 54.214 m + 135.299 m = 291.878 m.
To answer these questions, we can use the equations of motion, specifically the equations that relate velocity, time, and acceleration.
(a) To find the velocity of the boat at t = 21.7 s, we need to analyze the different time intervals and use the equations of motion for each interval.
In the first interval (t = 0 to t = 7.05 s), the boat starts from rest and accelerates at +2.01 m/s^2. To find the final velocity (v1) at the end of this interval, we can use the equation:
v1 = u + at1,
where u is the initial velocity (0 m/s as the boat starts from rest), a is the acceleration (+2.01 m/s^2), and t1 is the duration of this interval (7.05 s).
v1 = 0 + 2.01 * 7.05
In the second interval (t = 7.05 s to t = 13.2 s), the boat continues with a different acceleration of +0.518 m/s^2. Since the initial velocity at the start of this interval is the final velocity from the previous interval (v1), we can use the equation:
v2 = v1 + a2 * t2,
where a2 is the acceleration (+0.518 m/s^2) and t2 is the duration of this interval (6.15 s).
v2 = v1 + 0.518 * 6.15
In the third interval (t = 13.2 s to t = 21.7 s), the boat decelerates with an acceleration of -1.49 m/s^2. Again, since the initial velocity at the start of this interval is the final velocity from the previous interval (v2), we use the equation:
v3 = v2 + a3 * t3,
where a3 is the acceleration (-1.49 m/s^2) and t3 is the duration of this interval (8.5 s).
v3 = v2 + (-1.49) * 8.5
Finally, the velocity at t = 21.7 s would be v3.
(b) To find the total displacement of the boat, we can use the equations of motion and consider each interval separately.
In the first interval (t = 0 to t = 7.05 s), we can use the equation:
s1 = u * t1 + (1/2) * a * t1^2,
where s1 is the displacement, u is the initial velocity (0 m/s), a is the acceleration (+2.01 m/s^2), and t1 is the duration of this interval (7.05 s).
s1 = 0 * 7.05 + (1/2) * 2.01 * (7.05)^2
In the second interval (t = 7.05 s to t = 13.2 s), we can use the equation:
s2 = v1 * t2 + (1/2) * a2 * t2^2,
where v1 is the final velocity from the previous interval, a2 is the acceleration (+0.518 m/s^2), and t2 is the duration of this interval (6.15 s).
s2 = v1 * 6.15 + (1/2) * 0.518 * (6.15)^2
In the third interval (t = 13.2 s to t = 21.7 s), the boat decelerates; therefore, we need to use the equation for deceleration:
s3 = v2 * t3 + (1/2) * a3 * t3^2,
where v2 is the final velocity from the previous interval, a3 is the deceleration (-1.49 m/s^2), and t3 is the duration of this interval (8.5 s).
s3 = v2 * 8.5 + (1/2) * (-1.49) * (8.5)^2
Finally, the total displacement is the sum of s1, s2, and s3.