Question

image result for velocity vs. time graphs
which of the velocity vs. time graphs above represents an object that begins with a positive velocity, accelerates negatively, and finally ends up moving at approximately the same speed as it started, but in the opposite direction?
2
4
1
6
5
3

Explanation:

1. Understand the motion characteristics:

• The object starts with a positive velocity (so the graph starts above the time - axis).
• It accelerates negatively, which means the slope of the velocity - time graph is negative (the graph is a straight line or a curve with a negative slope).
• Finally, it ends up with a velocity that has the same magnitude as the initial velocity but opposite direction (so the final velocity is negative and has the same absolute value as the initial positive velocity).

1. Analyze the velocity - time graph features:

• For a velocity - time graph, the initial velocity \(v_0>0\), the acceleration \(a=\frac{\Delta v}{\Delta t}<0\) (negative slope), and the final velocity \(v = - v_0\) (same speed, opposite direction).
• A graph that starts at a positive velocity, has a negative slope (negative acceleration), and ends at a velocity with the same magnitude as the initial velocity but negative will be a straight line (if constant acceleration) or a curve with negative slope that goes from positive \(v\) to negative \(v\) with \(|v_{final}|=|v_{initial}|\).
• Among the given options (assuming the standard velocity - time graphs for such motion), the graph that satisfies these conditions is the one where the velocity starts positive, decreases (negative acceleration) until it becomes negative, and the magnitude of the final velocity is approximately equal to the magnitude of the initial velocity.

Answer:

(Assuming the correct option is the one that matches the described motion. Since the user has a marked option as 5, and based on the analysis above, if option 5's graph shows a positive initial velocity, negative slope (negative acceleration), and final velocity with same magnitude as initial but negative, then) 5. (The graph corresponding to option 5)