Question

question 4 given the following position vs time graph, what is happening to the object at c? position vs time graph options: the object is moving backwards with a constant velocity; the object is slowing down

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 represents the velocity of the object. A positive slope indicates motion in the positive direction (for example, forward motion if we consider the positive position direction as forward), and a constant slope means a constant velocity.

Looking at the graph, the line segment that contains point "C" has a constant slope. As time increases (moving to the right on the time axis), the position of the object is increasing (moving up on the position axis). But wait, let's re - evaluate the axes. Wait, the time axis here: wait, the vertical axis is time (sec) and the horizontal axis is position (m)? Wait, no, usually position - time graphs have time on the x - axis and position on the y - axis. But in this graph, it seems the vertical axis is time (sec) and horizontal is position (m). Wait, that's an unusual orientation, but the principle still holds: the slope of the line (change in position over change in time) gives velocity.

Wait, let's check the options. One option is "The object is moving backwards with a constant velocity" and the other is "The object is slowing down". Wait, no, let's re - analyze. Wait, if we consider the standard position - time graph (time on x, position on y), but here it's reversed. Let's take two points on the line containing C. Let's say at time t1 (vertical axis) and position x1 (horizontal axis), and another point t2 and x2. The slope would be (x2 - x1)/(t2 - t1). If as time (t) increases (moving down the vertical axis, since time is on the vertical), what happens to position (x)? Wait, the line from, say, the point at time 40 sec (vertical) and position 0 (horizontal) to time 0 sec and position 80 m? Wait, no, the blue line: let's look at the coordinates. Wait, the vertical axis is time (sec) from 0 to 50, and horizontal is position (m) from - 40 to 80. The blue line goes from (position = 0, time = 0) to (position = 60, time = 10), then to (position = 0, time = 40)? Wait, no, maybe I got the axes wrong.

Wait, maybe the horizontal axis is time (sec) and vertical is position (m). Let's re - interpret. If horizontal is time (t) and vertical is position (x). Then at point C, which is on a line segment. Let's take two points on that segment. Suppose at t = 0, x = 0; at t = 10, x = 60; at t = 20, x = 0? No, that doesn't make sense. Wait, the blue line: from (t = 0, x = 0) to (t = 10, x = 60) (point B), then from (t = 10, x = 60) to (t = 40, x = 0) (point D)? No, the line from B to D: let's calculate the slope. Slope = (x_D - x_B)/(t_D - t_B)=(0 - 60)/(40 - 10)= - 2 m/s. A negative slope means the object is moving in the negative direction (backwards) and since the slope is constant, the velocity is constant. So at point C, which is on this line segment, the object is moving backwards with a constant velocity. The other option "slowing down" would imply a changing slope (changing velocity), but the slope here is constant.

Answer:

The object is moving backwards with a constant velocity