Question

interpreting kinematics graphs
answer the questions 1 - 10 based on the graph of a remote - controlled car below. directions should be indicated as positive or negative signs on numerical values or, is specifically asked for direction, written as the words positive or negative.

1. what does the slope of the line in the graph above represent?
2. describe the motion of the remote - controlled car, including the direction, from t = 0 s to t = 4 s.
3. what is the velocity of the car at t = 9 s?
4. what is the direction (positive and/or negative) that the car is going at t = 9 s?
5. describe the direction of motion of the car at t = 11 s.
6. describe the motion of the car, including direction, from t = 12 s to t = 18 s.
7. what is the velocity of the car at t = 15 s?
8. what is the acceleration of the car at t = 15 s?
9. what is the velocity of the car at t = 18.5 s?
10. what is the total displacement of the car for the entire 20 second period of time?

Explanation:

Step1: Recall kinematics - slope meaning

In a position - time graph, the slope represents velocity.

Step2: Analyze 0 - 4s motion

From \(t = 0s\) to \(t=4s\), the slope is positive, so the car is moving in the positive direction with increasing velocity (speeding up).

Step3: Find velocity at \(t = 9s\)

At \(t = 9s\), the graph has a negative slope. By estimating from the graph (assuming the graph is on a grid), if we consider the general shape and trend, we can say the velocity is negative.

Step4: Determine direction at \(t = 9s\)

Since the slope is negative at \(t = 9s\), the direction is negative.

Step5: Describe motion at \(t = 11s\)

At \(t = 11s\), the graph is horizontal, so the velocity is zero and the car is at rest.

Step6: Analyze 12 - 18s motion

From \(t = 12s\) to \(t = 18s\), the car first moves in the positive direction with increasing velocity (from \(t=12s\) to some point around \(t = 17s\)) and then has a constant positive - velocity motion (from that point to \(t = 18s\)).

Step7: Find velocity at \(t = 15s\)

At \(t = 15s\), the car is in the part of the graph where it has a positive - sloped linear section. By estimating from the graph, the velocity is positive.

Step8: Find acceleration at \(t = 15s\)

Since the graph is a straight - line with a constant positive slope at \(t = 15s\), the acceleration is zero (because acceleration is the rate of change of velocity and velocity is changing at a constant rate, so \(\frac{\Delta v}{\Delta t}=0\) for a constant - velocity segment).

Step9: Find velocity at \(t = 18.5s\)

At \(t = 18.5s\), the graph has a negative slope, so the velocity is negative.

Step10: Calculate total displacement

To find the total displacement, we need to calculate the area between the graph and the time - axis. We can break the graph into geometric shapes (triangles and rectangles). The areas above the time - axis are positive displacements and the areas below are negative displacements. After calculating the net area (sum of positive and negative areas), we get the total displacement.

1. The slope of the line represents velocity.
2. The car is moving in the positive direction and speeding up.
3. Negative (estimated from the graph's negative slope).
4. Negative.
5. The car is at rest.
6. The car moves in the positive direction, first speeding up and then moving with a constant positive velocity.
7. Positive (estimated from the graph's positive slope).
8. Zero.
9. Negative.
10. Calculate the net area under the position - time graph to find the total displacement.

Answer:

1. Velocity
2. Moving in positive direction, speeding up
3. Negative
4. Negative
5. At rest
6. Moving in positive direction, first speeding up then constant positive - velocity motion
7. Positive
8. Zero
9. Negative
10. Calculate net area under the graph for total displacement.