https://file.aiquickdraw.com/mathbot/file/17302147649115sbuaus8.jpg

Let's solve each question in **Exercise 17.3** step-by-step.

### Question 1
Draw a distance-time graph for the first 10 seconds of an object traveling at \(6 \, \text{m/s}\).

1. Since the speed is constant at \(6 \, \text{m/s}\), the distance traveled after \(t\) seconds is given by \( \text{distance} = \text{speed} \times \text{time} \).
   
2. For the first 10 seconds:
   \[
   \text{distance} = 6 \times t
   \]
   So, for each second, we calculate:
   - \( t = 0 \): distance = \(6 \times 0 = 0 \, \text{m}\)
   - \( t = 1 \): distance = \(6 \times 1 = 6 \, \text{m}\)
   - \( t = 2 \): distance = \(6 \times 2 = 12 \, \text{m}\)
   - And so forth up to \( t = 10 \): distance = \(6 \times 10 = 60 \, \text{m}\).

3. The graph will be a straight line passing through these points, from \( (0, 0) \) to \( (10, 60) \).

### Question 2
Draw a distance-time graph for the first 10 seconds of an object traveling at \(5 \, \text{m/s}\). Use your graph to estimate:
   - (a) The time taken to travel \(25 \, \text{m}\).
   - (b) How far the object travels in \(3.5 \, \text{s}\).

1. Since the speed is \(5 \, \text{m/s}\), the distance after \(t\) seconds is given by:
   \[
   \text{distance} = 5 \times t
   \]
   - For \(t = 10\), \( \text{distance} = 5 \times 10 = 50 \, \text{m}\).

2. (a) To find the time to travel \(25 \, \text{m}\):
   \[
   t = \frac{\text{distance}}{\text{speed}} = \frac{25}{5} = 5 \, \text{s}
   \]

3. (b) To find the distance in \(3.5 \, \text{s}\):
   \[
   \text{distance} = 5 \times 3.5 = 17.5 \, \text{m}
   \]

### Question 3
Two objects \(A\) and \(B\) set off from the same point and move in the same straight line. \(B\) sets off first, while \(A\) sets off \(2\) seconds later. Use the graph to estimate:
   - (a) The speed of each object.
   - (b) How far apart the objects would be \(20\) seconds after the start.

1. From the graph:
   - \(B\) has a steeper line, indicating higher speed. Counting the grid lines:
     - \(B\) travels from \(0 \, \text{m}\) to \(70 \, \text{m}\) in \(8 \, \text{s}\), so:
       \[
       \text{speed of } B = \frac{70 \, \text{m}}{8 \, \text{s}} = 8.75 \, \text{m/s}
       \]
     - \(A\) travels from \(0 \, \text{m}\) to \(50 \, \text{m}\) in \(5 \, \text{s}\), so:
       \[
       \text{speed of } A = \frac{50 \, \text{m}}{5 \, \text{s}} = 10 \, \text{m/s}
       \]

2. (b) To find how far apart they are at \(20 \, \text{s}\), calculate the distance each would travel:
   - \(B\)'s distance = \(8.75 \times 20 = 175 \, \text{m}\).
   - \(A\)'s distance (starting at \(t = 2\)) = \(10 \times 18 = 180 \, \text{m}\).
   - Difference in distance: \(180 - 175 = 5 \, \text{m}\).

Would you like further details or have any questions about these steps?

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1. How is a distance-time graph useful for calculating speed?
2. What would happen to the graph if the speed of an object changed halfway?
3. How can we calculate the relative speed between two moving objects?
4. What does the slope of a line represent in a distance-time graph?
5. How does a delay in start time affect the graph of two moving objects?

**Tip:** When interpreting distance-time graphs, the steeper the slope, the faster the speed.

Solve Exercise 17.3 from the provided image, which involves drawing distance-time graphs and estimating speed, distance, and time for objects moving at constant speeds.

Explore how to solve distance-time graph problems for objects moving at constant speeds. Learn to calculate distance, speed, and time for various scenarios, such as estimating the distance traveled over a period or the separation between two objects.

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