Question

1. (10 points) the following graph shows an objects position vs time graph. the letters a to e correspond to various segments of the motion in which the graph has constant - slope.

a. in which segment(s) is the object at rest?

b. is the speed of the object during segment c greater than, equal to or less than its speed during segment e?

c. what is the objects position at the end of its trip?

d. what is the objects speed during segment a?

e. at what times does the object find itself at x = 0?

Explanation:

Step1: Identify constant - slope segments

On a position - time graph, constant slope means constant velocity. Segments with straight - line portions have constant slope. Segments A, B, C, and E have constant slopes.

Step2: Compare speeds

Speed is the magnitude of velocity. Velocity is the slope of the position - time graph. The slope of segment C is steeper (in magnitude) than the slope of segment E. So the speed in segment C is greater than the speed in segment E.

Step3: Determine final position

The object's position at the end of its trip is the value of x at the end of the graph. The object's position at the end of its trip is negative (at the end of segment E).

Step4: Calculate speed in segment A

The speed in segment A is the magnitude of the slope of segment A. The initial point of A is \((0,-1)\) and the end - point is \((2,0)\). The slope \(m=\frac{\Delta x}{\Delta t}=\frac{0 - (-1)}{2-0}=\frac{1}{2}\text{ m/s}\), so the speed is \(0.5\text{ m/s}\).

Step5: Find times when \(x = 0\)

The object is at \(x = 0\) when the graph intersects the \(t\) - axis (\(x = 0\)). The object is at \(x = 0\) at \(t = 2\text{ s}\) and \(t=14\text{ s}\).

Answer:

a. A, B, C, E
b. Greater
c. Negative
d. \(0.5\text{ m/s}\)
e. \(t = 2\text{ s}\) and \(t = 14\text{ s}\)