Final answer:
To describe the motion of a car from graphs, recognize that a straight slope in position vs. time indicates constant velocity, corresponding to a flat line on a velocity vs. time graph and zero on an acceleration vs. time graph. Curvature in the position graph indicates acceleration, reflected by changing slopes in the velocity graph and non-zero values in the acceleration graph.
Step-by-step explanation:
To describe the motion of a car based on various graphs, one must understand the relationship between position, velocity, and acceleration. The position vs. time graph displays how far the car is from the starting point at any given time. When this graph is a straight line with a constant slope, it indicates that the car is moving with a constant velocity. If the graph slopes upwards, the car is moving away from the start point; if it slopes downwards, it is returning.
A corresponding velocity vs. time graph would have a horizontal line if the car is moving with constant velocity, as the slope of the position vs. time graph (which represents velocity) is constant. The line on the velocity-time graph would be at the value of that constant velocity, above the time axis if moving forward, and below if reversing.
An acceleration vs. time graph would show a horizontal line at zero if the car is moving with a constant velocity since there is no change in velocity over time hence no acceleration.
If the position vs. time graph shows curvature, that indicates changing velocity – acceleration or deceleration. In this case, to create a velocity vs. time graph, you'd look at the steepness of the curve at various intervals. A steeper curve means higher velocity; the direction of the curve indicates whether the velocity is positive or negative. The acceleration vs. time graph would reflect changes in this slope or curve steepness over time. For example, a car that is speeding up will have a position curve that gets steeper and a positive acceleration, whereas a car that is slowing down will have a curve that flattens out and may have a negative acceleration if it's decelerating.