Question

16 opción múltiple 1 punto when is the object accelerating, according to the graph? from 0-2 seconds, 7-12 seconds, and 16-18 seconds from 2-5 seconds and 12-14 seconds never from 5-7 seconds, 14-16 seconds, and 18-19 seconds

Explanation:

To determine when an object is accelerating from a position - time graph, we use the fact that acceleration occurs when the velocity of the object is changing. Velocity is the slope of the position - time graph. So, we need to find the intervals where the slope of the position - time graph is changing (i.e., the graph is not a straight line segment, because a straight line segment has a constant slope, meaning constant velocity and zero acceleration).

Step 1: Analyze the slope of the position - time graph

- In a position - time graph, a straight line segment represents constant velocity (since velocity $v=\frac{\Delta x}{\Delta t}$, and for a straight line, $\frac{\Delta x}{\Delta t}$ is constant). When the graph has a non - straight (curved or changing slope) segment, the velocity is changing, and thus the object is accelerating. But in this graph, all the segments are straight lines. Wait, no, actually, the key is that acceleration is the rate of change of velocity. If the velocity is changing, the object is accelerating. Velocity is the slope of the position - time graph. So, when the slope is changing (i.e., when the graph is not a straight line, or when the direction of the slope changes abruptly, which is equivalent to a change in velocity), the object is accelerating.
- Let's analyze each interval:
- For the interval 0 - 2 seconds: The slope of the position - time graph is 0 (since the position does not change), so velocity $v = 0$ (constant), so acceleration $a=0$.
- For the interval 2 - 5 seconds: The graph is a straight line with a constant slope (negative slope), so velocity is constant, acceleration $a = 0$.
- For the interval 5 - 7 seconds: The slope of the graph changes (from a negative slope to a positive slope, and the graph is a straight line segment with a different slope than the previous segment). So the velocity is changing here, so the object is accelerating.
- For the interval 7 - 12 seconds: The graph is a straight line with a constant slope (zero slope? Wait, no, the position is constant? Wait, no, from 7 - 12 seconds, the position is constant at $y = 2$, so slope is 0, velocity is 0 (constant), acceleration $a=0$.
- For the interval 12 - 14 seconds: The graph is a straight line with a constant (negative) slope, so velocity is constant, acceleration $a = 0$.
- For the interval 14 - 16 seconds: The slope of the graph changes (from a negative slope to a more negative slope? Wait, no, the graph goes from $y = 2$ to $y=-6$ in a straight line? Wait, no, looking at the graph, from 12 - 14 seconds, the slope is negative, and from 14 - 16 seconds, the slope is more negative? Wait, no, actually, when the graph has a change in the slope (i.e., when the line changes its steepness or direction), the velocity changes. So from 5 - 7 seconds, the slope changes from negative to positive (the graph goes from a minimum at $t = 5$ to $y = 2$ at $t=7$), so velocity changes (from negative to positive), so acceleration occurs. From 14 - 16 seconds, the slope changes (the graph goes from a higher position to a lower position with a steeper slope), so velocity changes. From 18 - 19 seconds, the slope changes (the graph goes from $y=-2$ to $y = 0$), so velocity changes.
- Now let's check the options:
- Option 1: 0 - 2 (slope 0, constant velocity), 7 - 12 (slope 0, constant velocity), 16 - 18 (slope 0, constant velocity) → no acceleration.
- Option 2: 2 - 5 (constant slope, constant velocity), 12 - 14 (constant slope, constant velocity) → no acceleration.
- Option 3: "Never" is wrong because there are intervals where slope changes.
- Option 4: 5 - 7 seconds (slope changes from negative to positive, velocity changes), 14 - 16 seconds (slope changes, velocity changes), 18 - 19 seconds (slope changes, velocity changes) → acceleration occurs here.

Answer:

D. From 5 - 7 seconds, 14 - 16 seconds, and 18 - 19 seconds (assuming the last option is D, if the options are labeled as A, B, C, D with the last option being the one with 5 - 7, 14 - 16, 18 - 19)