Question

the graph shows the position of an object over time. position vs. time select from the drop - down menus to correctly complete the sentence. the velocity of the object is and the forces acting on the object are

Explanation:

To solve this, we analyze the position - time graph:

Step 1: Analyze the velocity

The velocity of an object is the slope of the position - time graph. For a position - time graph, the formula for the slope (which represents velocity) between two points \((t_1,x_1)\) and \((t_2,x_2)\) is \(v=\frac{\Delta x}{\Delta t}=\frac{x_2 - x_1}{t_2 - t_1}\). In a position - time graph, if the graph is a curve that is concave up (as in the given graph, the curve is increasing and getting steeper), the slope (velocity) is increasing. This is because as time \(t\) increases, the change in position \(\Delta x\) for a given change in time \(\Delta t\) is getting larger. So the velocity of the object is increasing.

Step 2: Analyze the forces

According to Newton's second law, \(F = ma\), where \(F\) is the net force, \(m\) is the mass of the object and \(a\) is the acceleration. Acceleration \(a=\frac{\Delta v}{\Delta t}\). Since the velocity \(v\) is increasing (from step 1), the acceleration \(a\) is positive (because \(\Delta v=v_2 - v_1>0\) for \(t_2>t_1\)). If \(a>0\) and \(m>0\) (mass of an object is always positive), then the net force \(F = ma>0\). Also, if the velocity is increasing, the acceleration is non - zero. For the velocity to be increasing, there must be a net force acting on the object. And if the velocity is increasing at an increasing rate (since the slope of the position - time graph is increasing, which means the rate of change of velocity (acceleration) is also increasing), the net force is also increasing. But in the context of a basic analysis (assuming a simple case where we can say that if velocity is increasing, the forces acting on the object are unbalanced (there is a net force) and in the direction of motion. If we consider the most basic case, when velocity is increasing, the net force is non - zero, so the forces acting on the object are unbalanced (not zero, and there is a net force).

So, the velocity of the object is increasing, and the forces acting on the object are unbalanced (or there is a net force acting on the object).

If we assume the drop - down options for velocity are something like "constant", "decreasing", "increasing" and for forces are "balanced" (net force = 0), "unbalanced" (net force\(
eq0\)):

• For the velocity part: The slope of the position - time graph is increasing (the graph is concave up), so the velocity is increasing.
• For the forces part: Since velocity is changing (increasing), according to Newton's second law \(F = ma\), if \(v\) is changing, \(a

eq0\), so \(F
eq0\), which means the forces are unbalanced.

So the completed sentence would be: "The velocity of the object is increasing and the forces acting on the object are unbalanced."

Answer:

To solve this, we analyze the position - time graph:

Step 1: Analyze the velocity

The velocity of an object is the slope of the position - time graph. For a position - time graph, the formula for the slope (which represents velocity) between two points \((t_1,x_1)\) and \((t_2,x_2)\) is \(v=\frac{\Delta x}{\Delta t}=\frac{x_2 - x_1}{t_2 - t_1}\). In a position - time graph, if the graph is a curve that is concave up (as in the given graph, the curve is increasing and getting steeper), the slope (velocity) is increasing. This is because as time \(t\) increases, the change in position \(\Delta x\) for a given change in time \(\Delta t\) is getting larger. So the velocity of the object is increasing.

Step 2: Analyze the forces

According to Newton's second law, \(F = ma\), where \(F\) is the net force, \(m\) is the mass of the object and \(a\) is the acceleration. Acceleration \(a=\frac{\Delta v}{\Delta t}\). Since the velocity \(v\) is increasing (from step 1), the acceleration \(a\) is positive (because \(\Delta v=v_2 - v_1>0\) for \(t_2>t_1\)). If \(a>0\) and \(m>0\) (mass of an object is always positive), then the net force \(F = ma>0\). Also, if the velocity is increasing, the acceleration is non - zero. For the velocity to be increasing, there must be a net force acting on the object. And if the velocity is increasing at an increasing rate (since the slope of the position - time graph is increasing, which means the rate of change of velocity (acceleration) is also increasing), the net force is also increasing. But in the context of a basic analysis (assuming a simple case where we can say that if velocity is increasing, the forces acting on the object are unbalanced (there is a net force) and in the direction of motion. If we consider the most basic case, when velocity is increasing, the net force is non - zero, so the forces acting on the object are unbalanced (not zero, and there is a net force).

So, the velocity of the object is increasing, and the forces acting on the object are unbalanced (or there is a net force acting on the object).

If we assume the drop - down options for velocity are something like "constant", "decreasing", "increasing" and for forces are "balanced" (net force = 0), "unbalanced" (net force\(
eq0\)):

• For the velocity part: The slope of the position - time graph is increasing (the graph is concave up), so the velocity is increasing.
• For the forces part: Since velocity is changing (increasing), according to Newton's second law \(F = ma\), if \(v\) is changing, \(a

eq0\), so \(F
eq0\), which means the forces are unbalanced.

So the completed sentence would be: "The velocity of the object is increasing and the forces acting on the object are unbalanced."