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answer the question 3,4, 5 and write the solution and formula

Let's solve each question step by step.

### Question 3:
**Problem:** An object moves from point A to point C along the rectangle shown in the figure on the right. Find the distance covered by the moving object.

### Solution:
To find the total distance covered by the object as it moves from point A to point C along the sides of the rectangle, we sum up the lengths of the two sides \(AB\) and \(BC\).

Given:
- Length of \(AB = 24 \, \text{km}\)
- Length of \(BC = 18 \, \text{km}\)

Total distance covered:
\[
\text{Distance} = AB + BC = 24 \, \text{km} + 18 \, \text{km} = 42 \, \text{km}
\]

**Answer:** The distance covered is **42 km**. The correct option is **B. 42 km**.

---

### Question 4:
**Problem:** Using the same figure, determine the magnitude of the displacement of the object.

### Solution:
Displacement is the straight-line distance between the starting point \(A\) and the ending point \(C\), which forms the hypotenuse of the right triangle \(ABC\).

We can use the Pythagorean theorem to calculate the displacement \(AC\):
\[
AC = \sqrt{AB^2 + BC^2}
\]

Substituting the given values:
\[
AC = \sqrt{(24 \, \text{km})^2 + (18 \, \text{km})^2}
\]
\[
AC = \sqrt{576 \, \text{km}^2 + 324 \, \text{km}^2}
\]
\[
AC = \sqrt{900 \, \text{km}^2} = 30 \, \text{km}
\]

**Answer:** The magnitude of the displacement is **30 km**. The correct option is **B. 30 km**.

---

### Question 5:
**Problem:** A train leaves with an average speed of \( 7.5 \, \text{m/s} \) only from the rest to another at 5.4 seconds. It moves before flying with an average speed of \( 5.0 \, \text{m/s} \) for 8.0s towards its other limits. What is the train’s total distance for its starting point?

### Solution:
The problem describes two stages of motion:

1. **First Stage:** 
   - Speed \(v_1 = 7.5 \, \text{m/s}\)
   - Time \(t_1 = 5.4 \, \text{s}\)

   Distance for the first stage:
   \[
   d_1 = v_1 \times t_1 = 7.5 \, \text{m/s} \times 5.4 \, \text{s} = 40.5 \, \text{m}
   \]

2. **Second Stage:** 
   - Speed \(v_2 = 5.0 \, \text{m/s}\)
   - Time \(t_2 = 8.0 \, \text{s}\)

   Distance for the second stage:
   \[
   d_2 = v_2 \times t_2 = 5.0 \, \text{m/s} \times 8.0 \, \text{s} = 40.0 \, \text{m}
   \]

Total distance traveled:
\[
\text{Total Distance} = d_1 + d_2 = 40.5 \, \text{m} + 40.0 \, \text{m} = 80.5 \, \text{m}
\]

**Answer:** The total distance is **80.5 m**. The correct option is **C. 80.5 m**.

---

Do you have any questions or need further details? Here are some related questions to consider:

1. How does the Pythagorean theorem help in finding the displacement in right-angle triangle scenarios?
2. What is the difference between distance and displacement?
3. How would the distance and displacement differ if the object moved along a curved path instead of a straight line?
4. How can you determine the total time taken if the average speed and total distance are given?
5. What is the significance of average speed in determining the distance traveled?

**Tip:** Remember that displacement is always the shortest path between the initial and final points, which is why it's calculated as a straight-line distance, unlike the total distance which accounts for the entire path taken.

Learn how to calculate distance and displacement with detailed step-by-step solutions. Includes problems involving rectangular paths and moving objects. Suitable for high school students.

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