Question

given the following position versus time graph, what is... the object at c? the object is moving with a constant positive velocity the object is moving with a positive acceleration

Explanation:

To determine the motion of the object at point \( C \) on the position - time graph, we use the concept that the slope of a position - time graph gives the velocity of the object, and the rate of change of the slope (the second derivative of position with respect to time, or the derivative of velocity with respect to time) gives the acceleration.

Step 1: Analyze the slope at point \( C \)

The slope of the position - time graph \( s(t) \) at a point is given by \( v=\frac{ds}{dt} \), which is the velocity. At point \( C \), we look at the tangent to the curve. The curve at \( C \) is concave down (since it is curving towards the time - axis). For a concave - down curve, the slope (velocity) is decreasing as time increases.

Step 2: Relate the change in slope to acceleration

Acceleration \( a=\frac{dv}{dt}=\frac{d^{2}s}{dt^{2}} \). If the slope (velocity) is decreasing, and velocity is positive (because the object is moving in the positive direction as the position is increasing), then the acceleration \( a=\frac{dv}{dt}<0 \) (negative acceleration, or deceleration). But let's re - evaluate the options. Wait, maybe we made a mistake. Wait, the options are about positive velocity or positive acceleration. Wait, the position - time graph: at point \( C \), the object is moving (since the slope is non - zero), and the slope is positive (because as time increases, position increases, so \( \frac{ds}{dt}>0 \), so velocity is positive). But what about acceleration? The curve is concave down, so the second derivative \( \frac{d^{2}s}{dt^{2}}<0 \), so acceleration is negative. But the options given: one says "The object is moving with a constant positive velocity" (which would be a straight line with constant slope, but at \( C \) the slope is changing, so not constant), and the other "The object is moving with a positive acceleration" (but we saw acceleration is negative). Wait, maybe the graph: let's re - interpret. Wait, the graph has three segments: \( A \) (a straight line with positive slope, constant velocity), \( B \) (a horizontal line, slope = 0, velocity = 0, object at rest), and \( C \) (a curve). At point \( C \), the tangent to the curve: as time increases, the slope of the tangent (velocity) is decreasing. But the velocity is still positive (because the object is moving in the positive direction, position is increasing). But the options: if we have to choose between the two (assuming these are the only two options, maybe the original question had more, but from the image, we can see two options). Wait, maybe the first option: "The object is moving with a constant positive velocity" is incorrect because the slope is changing (so velocity is not constant). The second option "The object is moving with a positive acceleration" is also incorrect? Wait, no, maybe I misread the graph. Wait, maybe the curve at \( C \) is concave up? Wait, the graph as shown: the curve at \( C \) is curving such that the slope is increasing? Wait, no, the position - time graph: if the curve is bending upwards (concave up), then the slope (velocity) is increasing, so acceleration is positive. If it is bending downwards (concave down), slope is decreasing, acceleration is negative. Wait, maybe the graph at \( C \) is concave up. Let's assume that. If the curve at \( C \) is concave up, then the slope (velocity) is increasing. Since the slope is positive (position increasing with time), an increasing positive slope means positive acceleration. But the first option: "The object is moving with a constant positive velocity" is wrong because the slope is changing. So the correct statement (if we assume the two options are the ones given) would be: Wait, maybe the question is asking which statement is true at \( C \). Let's re - express:

- For "The object is moving with a constant positive velocity": A position - time graph with constant positive velocity is a straight line with positive slope. At \( C \), the graph is a curve, so the slope (velocity) is changing. So this statement is false.
- For "The object is moving with a positive acceleration": If the curve is concave up (slope is increasing), then \( \frac{dv}{dt}=a > 0 \), so acceleration is positive. So this statement is true.

Wait, maybe the graph at \( C \) is concave up. Let's go back to the position - time graph interpretation. The slope of \( s(t) \) is \( v(t) \), and the slope of \( v(t) \) (the second derivative of \( s(t) \)) is \( a(t) \). If the \( s(t) \) curve is concave up, \( a(t)>0 \); if concave down, \( a(t)<0 \).

Assuming that at point \( C \), the curve is concave up (maybe the graph is drawn such that the curve is curving upwards), then the acceleration is positive, and the velocity is positive (since slope is positive). So the object is moving with positive acceleration.

Answer:

The object is moving with a positive acceleration