https://file.aiquickdraw.com/mathbot/file/17289883428632v4hhecn.jpg

Let's break down the problem for parts (a), (b), and (c):

### Problem Analysis:

1. **Tortoise**:
   - Speed: \( v_T = 0.5 \, \text{m/s} \) (constant)
   - Distance to cover: 60 meters
   - Time for the tortoise to finish the race: 
     \[
     t_T = \frac{\text{Distance}}{\text{Speed}} = \frac{60 \, \text{m}}{0.5 \, \text{m/s}} = 120 \, \text{seconds}
     \]

2. **Hare**:
   - The hare starts **100 seconds later**.
   - Accelerates uniformly for 10 seconds, reaching a top speed of \( 3 \, \text{m/s} \).
     - This means the acceleration \( a_H \) is given by:
       \[
       a_H = \frac{3 \, \text{m/s}}{10 \, \text{s}} = 0.3 \, \text{m/s}^2
       \]
     - Distance covered during acceleration: 
       \[
       d_{\text{acc}} = \frac{1}{2} a_H t^2 = \frac{1}{2} \times 0.3 \times (10)^2 = 15 \, \text{m}
       \]
   - Maintains top speed for 10 seconds:
     - Distance during this period: 
       \[
       d_{\text{constant}} = v_{\text{max}} \times t = 3 \times 10 = 30 \, \text{m}
       \]
   - Decelerates to rest in 10 seconds:
     - Deceleration: \( a_d = -\frac{3 \, \text{m/s}}{10 \, \text{s}} = -0.3 \, \text{m/s}^2 \)
     - Distance during deceleration: 
       \[
       d_{\text{dec}} = \frac{1}{2} v_{\text{max}} \times t = \frac{1}{2} \times 3 \times 10 = 15 \, \text{m}
       \]
   - Total distance covered by the hare: 
     \[
     d_{\text{hare}} = 15 + 30 + 15 = 60 \, \text{m}
     \]
   - Time taken for the hare to finish the race:
     \[
     t_H = 10 + 10 + 10 = 30 \, \text{seconds}
     \]

### (a) Plotting the Velocity-Time Graph:

1. **Tortoise**: 
   - Constant velocity of \( 0.5 \, \text{m/s} \) for 120 seconds.
   
2. **Hare**:
   - Accelerates uniformly from 0 to \( 3 \, \text{m/s} \) over 10 seconds.
   - Maintains speed of \( 3 \, \text{m/s} \) for 10 seconds.
   - Decelerates uniformly back to 0 over 10 seconds.
   - Starts 100 seconds after the tortoise.

### (b) Who Wins the Race?

- The tortoise takes 120 seconds to finish.
- The hare starts after 100 seconds and finishes in 30 seconds. Thus, the hare finishes at \( 100 + 30 = 130 \, \text{seconds} \).

Therefore, **the tortoise wins**, finishing 10 seconds earlier than the hare.

### (c) Time Between Crossings:

- The tortoise finishes in 120 seconds.
- The hare finishes in 130 seconds.
  
So, the time between the two animals crossing the finish line is:
\[
\Delta t = 130 - 120 = 10 \, \text{seconds}
\]

---

Would you like me to sketch the velocity-time graph for you? Do you have any questions about the calculations?

Here are 5 related questions you might consider:

1. How does the total distance traveled change if the hare starts accelerating sooner?
2. What would happen if the tortoise had a speed boost midway through the race?
3. If the race was 100 meters long instead, how would the results change?
4. What would happen if the hare's acceleration was faster or slower?
5. How can the time of the hare be minimized to finish the race faster?

**Tip:** Always compare the **areas under velocity-time graphs** when finding distances traveled by different objects.

A tortoise and a hare are running a race of 60 meters. The tortoise starts with a speed of 0.5 m/s, while the hare starts 100 seconds later, accelerating to 3 m/s in 10 seconds, maintaining that speed for another 10 seconds, and then decelerating to rest over 10 seconds. Plot velocity-time graph, determine who wins, and calculate the time difference between their finishes.

This problem involves a tortoise moving at a constant speed of 0.5 m/s and a hare that accelerates, maintains speed, and decelerates over a race distance of 60 meters. The solution involves plotting velocity-time graphs, calculating who wins the race, and finding the time difference between their finishes.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation