Question

version -01 - unit 2: quest 2: position - time graphs
this print - out should have 12 questions. multiple - choice questions may continue on the next column or page - find all choices before answering.
001 (part 1 of 2) 10.0 points
displacement curve 02 97775
consider a moving object whose position x is plotted as a function of the time t. the object moved in different ways during the time intervals denoted i, ii and iii on the figure.
during these three intervals, when was the object’s speed highest? do not confuse the speed with the velocity.

1. during interval i
2. during interval ii
3. during interval iii
4. same speed during intervals ii and iii
5. same speed during each of the three intervals.

002 (part 2 of 2) 10.0 points
during which interval(s) did the object’s velocity remain constant?

1. during interval i only
2. during interval ii only
3. during interval iii only
4. during each of the three intervals
5. during none of the three intervals

003 (part 1 of 3) 10.0 points
position vs time 05 206786
the scale on the horizontal axis is 3 s per division and on the vertical axis 4 m per division.
what is the initial position?
004 (part 2 of 3) 10.0 points
what is the final position?
answer in units of m. answer in units of m.
005 (part 3 of 3) 10.0 points
what velocity is represented by the graph?
answer in units of m/s. answer in units of m/s.
006 (part 1 of 7) 10.0 points
graphical analysis 01 192093
consider the following graph of motion.
how many meters can swimmer 1 cover in 30 seconds?

1. 10 m
2. 20 m

Explanation:

Step1: Identify initial position for 003

For the position - time graph with horizontal - axis scale of 3 s per division and vertical - axis scale of 4 m per division, at $t = 0$ (the start), the position is at the origin of the graph. So the initial position $x_0=0\times4 = 0$ m.

Step2: Identify final position for 004

At the end of the time - interval shown on the graph (last point on the curve), the vertical position is at 6 divisions on the vertical axis. So the final position $x_f=6\times4 = 24$ m.

Step3: Calculate velocity for 005

The velocity $v=\frac{\Delta x}{\Delta t}$. The change in position $\Delta x = 24 - 0=24$ m, and the change in time $\Delta t=8\times3 = 24$ s. So $v=\frac{24}{24}=1$ m/s.

Answer:

003: 0 m
004: 24 m
005: 1 m/s