https://file.aiquickdraw.com/mathbot/file/1727112289160i1lew2nk.jpg

To solve the given problem, we need to determine the ratio of the cart's displacement from \( t = 0 \) to \( t = 2 \) to the displacement from \( t = 2 \) to \( t = 4 \). Let's break it down step by step using the velocity-time graph.

### Step 1: Displacement from \( t = 0 \) to \( t = 2 \)
Displacement is the area under the velocity-time graph.

- From \( t = 0 \) to \( t = 1 \): The velocity decreases from 2 m/s to 1 m/s, forming a trapezoid.
  \[
  \text{Area} = \frac{1}{2} \times (2 + 1) \times 1 = \frac{3}{2} \, \text{m}
  \]

- From \( t = 1 \) to \( t = 2 \): The velocity decreases from 1 m/s to 0 m/s, forming a triangle.
  \[
  \text{Area} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \, \text{m}
  \]

Thus, the total displacement from \( t = 0 \) to \( t = 2 \) is:
\[
\text{Total Displacement (0 to 2)} = \frac{3}{2} + \frac{1}{2} = 2 \, \text{m}
\]

### Step 2: Displacement from \( t = 2 \) to \( t = 4 \)
Again, we calculate the area under the graph.

- From \( t = 2 \) to \( t = 3 \): The velocity increases from 0 m/s to 2 m/s, forming a triangle.
  \[
  \text{Area} = \frac{1}{2} \times 2 \times 1 = 1 \, \text{m}
  \]

- From \( t = 3 \) to \( t = 4 \): The velocity increases from 2 m/s to 4 m/s, forming another triangle.
  \[
  \text{Area} = \frac{1}{2} \times 4 \times 1 = 2 \, \text{m}
  \]

Thus, the total displacement from \( t = 2 \) to \( t = 4 \) is:
\[
\text{Total Displacement (2 to 4)} = 1 + 2 = 3 \, \text{m}
\]

### Step 3: Ratio of the Displacements
The ratio of the magnitude of the displacement from \( t = 0 \) to \( t = 2 \) to the displacement from \( t = 2 \) to \( t = 4 \) is:
\[
\frac{\text{Displacement (0 to 2)}}{\text{Displacement (2 to 4)}} = \frac{2}{3}
\]

Thus, the ratio is \( \boxed{\frac{2}{3}} \).

---

Would you like more details on any part of this? Here are some related questions to expand on this topic:

1. How do you calculate displacement from a velocity-time graph in general?
2. What is the difference between displacement and distance in physics?
3. How can you use integration to calculate displacement for non-linear velocity functions?
4. How does the shape of the velocity-time graph affect the motion of an object?
5. Can you calculate the total distance traveled, considering the cart’s change in direction?

**Tip:** Always remember that the area under a velocity-time graph gives displacement, and you should account for whether the area is above or below the time axis.

A cart moves along a straight line. The cart’s velocity as a function of time is shown in the graph. At t = 2s the cart’s motion changes. Based on the graph, what is the ratio of the magnitude of the cart’s displacement from t = 0 to t = 2s to the cart’s displacement from t = 2s to t = 4s?

Learn how to calculate the ratio of a cart's displacement using a velocity-time graph. This problem involves calculating the area under the graph to find displacement and includes key concepts in kinematics. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation