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.
V1 = 1.69m/s^2 * 7.18s = 12.13m/s.
V2 = 0.468m/s^2 * 4.55s = 2.13m/s.
V3 = -0.657m/s^2 * 8.38s = -5.51m/s.
a. V = V1 + V2 + V3 = 12.13 + 2.13 - 5.51 = 8.75m/s at 20.11s.
i got the right answer right for acceleration ... to find the total displacement i did 8.75 times 20.11 but its not the right answer
To solve this problem, we need to calculate the velocity of the boat at t = 20.11 s and the total displacement of the boat. We can break down the problem into four intervals based on the given acceleration values and durations.
Let's go step by step:
Interval 1: From t = 0 to t = 7.18 s
The boat starts from rest, so the initial velocity (u) is 0 m/s. The acceleration (a) is +1.69 m/s², and the duration (t) is 7.18 s.
We can use the kinematic equation: v = u + at, where v is the final velocity.
Plugging in the values, we have v1 = 0 + (1.69 m/s²)(7.18 s) = 12.14 m/s.
Interval 2: From t = 7.18 s to t = 11.73 s (7.18 s + 4.55 s)
The boat continues with an acceleration (a) of +0.468 m/s² for a duration (t) of 4.55 s.
Now, the initial velocity (u) is the final velocity (v1) from the previous interval, which is 12.14 m/s.
Using the same kinematic equation, we can find v2 = 12.14 m/s + (0.468 m/s²)(4.55 s) = 14.36 m/s.
Interval 3: From t = 11.73 s to t = 20.11 s
The boat accelerates at -0.657 m/s² for a duration of 8.38 s.
For this interval, we have negative acceleration, so we need to be careful with the signs.
The initial velocity (u) is the final velocity (v2) from the previous interval, which is 14.36 m/s.
Using the kinematic equation, v = u + at, we find v3 = 14.36 m/s + (-0.657 m/s²)(8.38 s) = 8.32 m/s.
Interval 4: From t = 20.11 s to t = 20.11 s
Since there is no acceleration given at this interval, the velocity remains constant.
The final velocity (v4) is the same as the previous interval, which is 8.32 m/s.
(a) The velocity of the boat at t = 20.11 s is 8.32 m/s.
To calculate the total displacement of the boat, we need to find the sum of the individual displacements in each interval.
Displacement in Interval 1:
Using the equation of motion, s = ut + (1/2)at², we find s1 = (0)(7.18 s) + (1/2)(1.69 m/s²)(7.18 s)² = 54.78 m.
Displacement in Interval 2:
The initial velocity (u) is 12.14 m/s, and the acceleration (a) is 0.468 m/s² for 4.55 s.
Using the same equation, s2 = (12.14 m/s)(4.55 s) + (1/2)(0.468 m/s²)(4.55 s)² = 30.82 m.
Displacement in Interval 3:
The initial velocity (u) is 14.36 m/s, and the acceleration (a) is -0.657 m/s² for 8.38 s.
Using the equation, s = ut + (1/2)at², we find s3 = (14.36 m/s)(8.38 s) + (1/2)(-0.657 m/s²)(8.38 s)² = 70.28 m.
Displacement in Interval 4:
Since there is no acceleration, the displacement in this interval is v4 multiplied by the duration, which is (8.32 m/s)(0) = 0 m.
(b) The total displacement of the boat is the sum of the individual displacements:
Total displacement = s1 + s2 + s3 + s4 = 54.78 m + 30.82 m + 70.28 m + 0 m = 155.88 m.
Thus, the total displacement of the boat is 155.88 m.