Given the graph below with the initial velocity, vi = -8 m/s, the object starts with one value of constant acceleration and stops at 2.3 seconds. Then, the value of its acceleration changes to a different acceleration. Finally, at t1 = 6.8 seconds, it travels at a constant (final) velocity. After t2 = 10.3 seconds, the average velocity of the entire trip is 3.8 m/s in the positive x direction. Determine the final velocity in m/s.
hw2_3.JPG
To determine the final velocity of the object in m/s, we need to analyze the given information step-by-step.
Step 1: Find the displacement during the first motion.
From the graph, we can see that the first motion lasts for 2.3 seconds. The area under the velocity-time graph represents displacement.
Displacement during the first motion = area under the velocity-time graph = (base) * (height) = (2.3 s) * (-8 m/s) = -18.4 m
Step 2: Find the displacement during the second motion.
From the information given, we know that the object changes its acceleration after 2.3 seconds. The velocity becomes constant after t1 = 6.8 seconds.
Displacement during the second motion = total displacement - displacement during the first motion
= average velocity * time interval - displacement during the first motion
= (3.8 m/s) * (10.3 s - 6.8 s) - (-18.4 m)
= (3.8 m/s) * (3.5 s) + 18.4 m
= 13.3 m + 18.4 m
= 31.7 m
Step 3: Find the final velocity.
From the graph, we know that the object reaches its final velocity after t2 = 10.3 seconds.
Final velocity = average velocity during the second motion = (total displacement) / (time interval)
= (31.7 m) / (10.3 s)
≈ 3.08 m/s
Therefore, the final velocity of the object is approximately 3.08 m/s.
To determine the final velocity of the object, we need to break down the motion into different segments and calculate the velocity for each segment. Then, we can add up the velocities to get the final velocity.
Segment 1: Initial velocity (vi) to first change in acceleration
Given data: vi = -8 m/s, t1 = 2.3 s
Using the equation of motion: vf = vi + at, we can find the velocity after t1:
vf1 = -8 + a(t1)
Segment 2: First change in acceleration to second change in acceleration
Given data: t1 = 2.3 s, t2 = 6.8 s
Since we know the initial velocity in this segment is the final velocity from the previous segment, we can calculate the displacement during this segment using the equation: Δx = vi*t + 0.5*a*t^2
Substituting the known values, the displacement during this segment is:
Δx2 = vf1(t2 - t1) + 0.5*a2(t2 - t1)^2
Segment 3: Second change in acceleration to final velocity
Given data: t2 = 6.8 s, tf = 10.3 s
Similarly to segment 2, we can calculate the displacement during this segment:
Δx3 = vf2(tf - t2) + 0.5*a3(tf - t2)^2
The total displacement for the trip is the sum of the displacements in each segment:
Δx_total = Δx2 + Δx3
We also know that the average velocity over the entire trip is 3.8 m/s. The average velocity is given by:
average velocity = total displacement / total time
3.8 = Δx_total / (tf - t1)
Now, we have two equations:
1. vf1 = -8 + a(t1)
2. 3.8 = Δx_total / (tf - t1)
We need one more equation to solve for the final velocity. Since we have constant acceleration in each segment, we can use the equation:
vf^2 = vi^2 + 2ad
We can solve for the final velocities in each segment using this equation:
1. vf1^2 = vi^2 + 2a1d1
2. vf2^2 = vf1^2 + 2a2d2
where d1 and d2 are the displacements in segments 2 and 3, respectively.
Using these equations, we can solve for vf1 and vf2, and then find the final velocity (vf) by adding up the velocities in each segment:
vf = vf1 + vf2