To solve this problem, we need to understand how velocity-time graphs work.

1. The car starts from rest and accelerates for 9 seconds to reach 16 m/s. This means the car is increasing its speed at a constant rate. On a velocity-time graph, this is represented by a straight line that slopes upwards. The slope of the line represents the acceleration. The steeper the slope, the greater the acceleration.

2. The car then travels at a constant velocity of 16 m/s for 4 seconds. On a velocity-time graph, constant velocity is represented by a horizontal line. This line would be at the 16 m/s mark on the y-axis and extend for 4 seconds on the x-axis.

3. Finally, the car decelerates at a constant rate to decrease its velocity by 4 m/s over 6 seconds. This means the car is slowing down at a constant rate. On a velocity-time graph, this is represented by a straight line that slopes downwards. The slope of the line represents the deceleration. The steeper the slope, the greater the deceleration. This line would start at the end of the 4 second constant velocity line and slope downwards to the 12 m/s mark on the y-axis (16 m/s - 4 m/s = 12 m/s) over 6 seconds on the x-axis.

Without seeing the actual lines A-F, I can't tell you which one completes the graph. However, the line should start at 0, slope upwards to 16 m/s at 9 seconds, stay horizontal at 16 m/s for 4 seconds, and then slope downwards to 12 m/s over the next 6 seconds.

Question

To solve this problem, we need to understand how velocity-time graphs work.

1. The car starts from rest and accelerates for 9 seconds to reach 16 m/s. This means the car is increasing its speed at a constant rate. On a velocity-time graph, this is represented by a straight line that slopes upwards. The slope of the line represents the acceleration. The steeper the slope, the greater the acceleration.

2. The car then travels at a constant velocity of 16 m/s for 4 seconds. On a velocity-time graph, constant velocity is represented by a horizontal line. This line would be at the 16 m/s mark on the y-axis and extend for 4 seconds on the x-axis.

3. Finally, the car decelerates at a constant rate to decrease its velocity by 4 m/s over 6 seconds. This means the car is slowing down at a constant rate. On a velocity-time graph, this is represented by a straight line that slopes downwards. The slope of the line represents the deceleration. The steeper the slope, the greater the deceleration. This line would start at the end of the 4 second constant velocity line and slope downwards to the 12 m/s mark on the y-axis (16 m/s - 4 m/s = 12 m/s) over 6 seconds on the x-axis.

Without seeing the actual lines A-F, I can't tell you which one completes the graph. However, the line should start at 0, slope upwards to 16 m/s at 9 seconds, stay horizontal at 16 m/s for 4 seconds, and then slope downwards to 12 m/s over the next 6 seconds.

Knowee AI · Accepted Answer

To solve this problem, we need to understand how velocity-time graphs work.

1. The car starts from rest and accelerates for 9 seconds to reach 16 m/s. This means the car is increasing its speed at a constant rate. On a velocity-time graph, this is represented by a straight line that slopes upwards. The slope of the line represents the acceleration. The steeper the slope, the greater the acceleration.

2. The car then travels at a constant velocity of 16 m/s for 4 seconds. On a velocity-time graph, constant velocity is represented by a horizontal line. This line would be at the 16 m/s mark on the y-axis and extend for 4 seconds on the x-axis.

3. Finally, the car decelerates at a constant rate to decrease its velocity by 4 m/s over 6 seconds. This means the car is slowing down at a constant rate. On a velocity-time graph, this is represented by a straight line that slopes downwards. The slope of the line represents the deceleration. The steeper the slope, the greater the deceleration. This line would start at the end of the 4 second constant velocity line and slope downwards to the 12 m/s mark on the y-axis (16 m/s - 4 m/s = 12 m/s) over 6 seconds on the x-axis.

Without seeing the actual lines A-F, I can't tell you which one completes the graph. However, the line should start at 0, slope upwards to 16 m/s at 9 seconds, stay horizontal at 16 m/s for 4 seconds, and then slope downwards to 12 m/s over the next 6 seconds.

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