https://file.aiquickdraw.com/mathbot/file/1730295816955py5k6tzx.jpg

Let's break down the given problem and analyze the information provided in the acceleration-time graph.

### Problem Analysis:
1. **Graph Description**: The acceleration-time graph shows that:
   - From \( t = 0 \) to \( t = 3 \) seconds, the car has a constant acceleration of \( 2 \, \text{m/s}^2 \).
   - From \( t = 3 \) to \( t = 5 \) seconds, the car’s acceleration linearly increases from \( 2 \, \text{m/s}^2 \) to \( 5 \, \text{m/s}^2 \).
   - From \( t = 5 \) to \( t = 6 \) seconds, the acceleration decreases sharply from \( 5 \, \text{m/s}^2 \) to \( -5 \, \text{m/s}^2 \).
   - From \( t = 6 \) to \( t = 7 \) seconds, the car maintains a constant acceleration of \( -5 \, \text{m/s}^2 \).

2. **Initial Speed**: When the car passes point \( P \) (at \( t = 0 \)), its speed is \( 12 \, \text{m/s} \).

3. **Questions**:
   - **(i)** Explain why the speed is greatest at \( t = 6 \) seconds.
   - **(ii)** Calculate the speed at \( t = 3 \) seconds and the distance traveled from \( P \) up to this time.

---

### Solution:

#### Part (i) Explanation:
The speed is the greatest at \( t = 6 \) seconds because the car has been accelerating in the positive direction up until this point, increasing its velocity. At \( t = 6 \) seconds, the car reaches the maximum positive acceleration (the peak area under the curve), after which the acceleration becomes negative, causing the speed to decrease. Thus, \( t = 6 \) is the point of maximum speed.

#### Part (ii) Calculations:

1. **Speed at \( t = 3 \) seconds**:
   - From \( t = 0 \) to \( t = 3 \), the acceleration is constant at \( 2 \, \text{m/s}^2 \).
   - Using the formula \( v = u + at \), where:
     - \( u = 12 \, \text{m/s} \) (initial speed),
     - \( a = 2 \, \text{m/s}^2 \) (acceleration),
     - \( t = 3 \) seconds.
   
   \[
   v = 12 + (2 \times 3) = 12 + 6 = 18 \, \text{m/s}
   \]
   So, the speed at \( t = 3 \) seconds is \( 18 \, \text{m/s} \).

2. **Distance from \( P \) at \( t = 3 \) seconds**:
   - Using the formula \( s = ut + \frac{1}{2}at^2 \), where:
     - \( u = 12 \, \text{m/s} \),
     - \( a = 2 \, \text{m/s}^2 \),
     - \( t = 3 \) seconds.

   \[
   s = (12 \times 3) + \frac{1}{2} \times 2 \times (3)^2
   \]
   \[
   s = 36 + \frac{1}{2} \times 2 \times 9
   \]
   \[
   s = 36 + 9 = 45 \, \text{meters}
   \]

   So, the distance from \( P \) at \( t = 3 \) seconds is \( 45 \, \text{meters} \).

---

Would you like further details on any step or have additional questions?

### Related Questions:
1. What is the speed of the car at \( t = 6 \) seconds?
2. How would the graph look if the car decelerated from the start?
3. What is the total distance traveled by the car in the 7 seconds shown on the graph?
4. How does varying acceleration affect the total distance traveled?
5. How would the velocity-time graph look for the same scenario?

### Tip:
When working with acceleration-time graphs, remember that the area under the curve represents the change in velocity over time.

A car is travelling due east along a straight road when it passes a point P. The acceleration of the car during the next 7 seconds is modeled in the acceleration-time graph (as shown in the provided image), where a in m/s² is the acceleration of the car due east, and t seconds is the time after passing the point P.

(i) Explain why the speed of the car is greatest when t = 6.
(ii) Calculate the speed of the car when t = 3 and the distance of the car from P at this time.

Learn how to solve a kinematics problem involving an acceleration-time graph, where we calculate the speed and distance of a car traveling eastward. This includes using key kinematic formulas to determine the car's speed at specific times and to explain maximum speed. 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