Question

1. position time velocity time acceleration time vel: 0 m acc: description: 4. position time velocity time acceleration time vel: 0 m acc: description:

Explanation:

For the first set of graphs (left - hand side):

• The position - time graph is a curve. The slope of the position - time graph gives the velocity. Since the slope is positive and decreasing (the curve is getting less steep as time increases), the velocity is positive but decreasing.
• The velocity - time graph is a horizontal line above the time - axis (positive velocity). But wait, there is a contradiction. However, if we consider the acceleration - time graph (horizontal line, constant acceleration). Using the equations of motion \(v = v_0+at\) and \(x=x_0 + v_0t+\frac{1}{2}at^{2}\). If \(a<0\) (negative acceleration), \(v = v_0+at\) (where \(v_0>0\)) will decrease over time. The position \(x=x_0 + v_0t+\frac{1}{2}at^{2}\) (a quadratic function with \(a<0\) for acceleration) will have a curve that is concave - down.
• Description: An object moving in the positive direction with a constant negative acceleration. Its initial velocity is positive, and as time passes, its velocity decreases (but remains positive for some time), and its position still increases but at a decreasing rate.

For the second set of graphs (right - hand side):

• Before the dashed line:
• The position - time graph is a horizontal line (slope = 0), so velocity \(v = 0\).
• The velocity - time graph is a horizontal line on the time - axis (\(v = 0\)).
• The acceleration - time graph is a horizontal line (constant acceleration, but since \(v = 0\) and not changing, \(a = 0\)).
• After the dashed line:
• The position - time graph has a negative slope (velocity is negative).
• The velocity - time graph is a horizontal line below the time - axis (negative velocity).
• The acceleration - time graph is a horizontal line (constant acceleration). If we assume a non - zero acceleration at the dashed line (a sudden change in velocity). But if we consider the overall (assuming the acceleration is non - zero only at the instant of the direction change, but the graph shows a constant acceleration. A more physical interpretation:
• Description: An object is at rest (velocity \(v = 0\), position constant) for some time. Then, it starts moving in the negative direction with a constant negative velocity (which implies zero acceleration for the motion in the negative direction part. But if we consider the acceleration graph is constant, we can say: An object is initially at rest. Then, it experiences a constant negative acceleration (which causes it to move in the negative direction with a velocity that changes at a constant rate. Mathematically, if \(v=v_0 + at\) and \(v_0 = 0\), \(v=at\) (negative if \(a<0\)) and \(x=x_0+\frac{1}{2}at^{2}\) (a quadratic function opening downwards for \(a<0\))

Answer:

Left - hand side description: An object moving in the positive direction with a constant negative acceleration. Its initial velocity is positive, and as time passes, its velocity decreases (but remains positive for some time), and its position still increases but at a decreasing rate.

Right - hand side description: An object is initially at rest. Then, it experiences a constant negative acceleration (which causes it to move in the negative direction with a velocity that changes at a constant rate.