https://file.aiquickdraw.com/mathbot/file/1729786292530gumzkcpp.jpg

Let's break down the problems from the image:

**Question 1.1** Define velocity.
- **Answer:** Velocity is the rate of change of displacement of an object with respect to time. It is a vector quantity, meaning it has both magnitude (speed) and direction. Mathematically, it's expressed as \( v = \frac{\Delta x}{\Delta t} \), where \( \Delta x \) is the change in position (displacement) and \( \Delta t \) is the change in time.

**Question 1.2** Calculate the magnitude of the acceleration of Car A in the first 5 seconds.
- From the graph, Car A starts from rest and follows a curved path, indicating non-uniform motion, meaning it is accelerating.
  - At \( t = 0 \), the position of Car A is \( 0 \, \text{m} \).
  - At \( t = 5 \, \text{s} \), the position of Car A is \( 25 \, \text{m} \).
  
  Since the motion is uniformly accelerated, we use the formula for displacement under constant acceleration:
  \[
  x = ut + \frac{1}{2} a t^2
  \]
  Here, \( u = 0 \, \text{m/s} \) (starts from rest), \( x = 25 \, \text{m} \), and \( t = 5 \, \text{s} \). Solving for acceleration \( a \):
  \[
  25 = 0 \times 5 + \frac{1}{2} a \times 5^2
  \]
  \[
  25 = \frac{1}{2} a \times 25
  \]
  \[
  a = 2 \, \text{m/s}^2
  \]
  So, the magnitude of acceleration is \( 2 \, \text{m/s}^2 \).

**Question 1.3** Compare the positions of the cars at 5 seconds.
- At \( t = 5 \, \text{s} \), Car A has reached 25 meters, as seen from the graph.
- Car B only starts moving at \( t_1 = 5 \, \text{s} \), so its position is \( 0 \, \text{m} \) at \( t = 5 \, \text{s} \).

**Question 1.4** Calculate the speed that Car A reaches after 5 seconds.
- The speed of Car A can be found using the final velocity formula for uniformly accelerated motion:
  \[
  v = u + at
  \]
  Given \( u = 0 \), \( a = 2 \, \text{m/s}^2 \), and \( t = 5 \, \text{s} \):
  \[
  v = 0 + 2 \times 5 = 10 \, \text{m/s}
  \]
  So, Car A reaches a speed of \( 10 \, \text{m/s} \) after 5 seconds.

**Question 1.5** Calculate the speed of Car B between 5 and 9 seconds.
- From the graph, Car B starts at \( t_1 = 5 \, \text{s} \) and moves linearly, indicating constant velocity.
- At \( t = 9 \, \text{s} \), Car B reaches a position of \( 85 \, \text{m} \).
- Therefore, the velocity of Car B is:
  \[
  v = \frac{\Delta x}{\Delta t} = \frac{85 - 0}{9 - 5} = \frac{85}{4} = 21.25 \, \text{m/s}
  \]
  So, the speed of Car B is \( 21.25 \, \text{m/s} \).

Let me know if you'd like further details on any of these answers, or need help with the remaining parts of the problem.

Here are 5 questions to further explore this topic:
1. What is the relationship between velocity and displacement?
2. How can you calculate distance traveled if acceleration is not constant?
3. What conditions must be true for an object to have zero velocity but non-zero acceleration?
4. How would the graph change if Car B experienced acceleration instead of constant velocity?
5. Can an object have a negative acceleration and still increase its speed?

**Tip:** When solving motion problems, always identify known variables (initial speed, time, displacement) and pick equations of motion accordingly.

A physics problem involving the motion of two cars (A and B) with questions related to velocity, acceleration, and displacement.

Solve a kinematics problem involving two cars. Learn to define velocity, calculate acceleration, and compare positions of the cars after a specific time. The solution uses key kinematic formulas and principles, suitable for high school students.

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