Question

complex velocity - time graphs
practice

1. for each segment determine the displacement. assuming the object starts from home, what is the objects final position?

a 100 ms
b 300 ms
c 75 ms
\\( \vec { d } _ { \text { final } } = 475 m s \\) of home

Explanation:

Step1: Calculate displacement for segment A

The formula for displacement from a velocity - time graph is \(d=\frac{1}{2}(v_{i} + v_{f})t\). For segment A, \(v_{i}=10\ m/s\), \(v_{f}=30\ m/s\), and \(t = 5\ s\).

$$d_{A}=\frac{1}{2}(10 + 30)\times5=\frac{1}{2}\times40\times5 = 100\ m$$

Step2: Calculate displacement for segment B

For segment B (constant velocity), \(v = 30\ m/s\) and \(t=(15 - 5)=10\ s\). Using \(d=v\times t\)

$$d_{B}=30\times10 = 300\ m$$

Step3: Calculate displacement for segment C

For segment C, \(v_{i}=30\ m/s\), \(v_{f}=0\ m/s\), and \(t=(20 - 15)=5\ s\). Using \(d=\frac{1}{2}(v_{i}+v_{f})t\)

$$d_{C}=\frac{1}{2}(30 + 0)\times5=\frac{1}{2}\times30\times5=75\ m$$

Step4: Calculate final position

Since the object starts from home (\(d_{initial}=0\)), the final position \(d_{final}=d_{A}+d_{B}+d_{C}\)

$$d_{final}=100 + 300+75=475\ m$$

Answer:

The displacement for segment A is \(100\ m\), for segment B is \(300\ m\), for segment C is \(75\ m\), and the object's final position is \(475\ m\) from home.