question 5 1 pts based upon the velocity - time graph below, the position at c is
velocity (m/s)
6
5
4
3
2
1
0
-1
-2
-3
-4
2
4
6
8
10
a
c
b
has a horizontal slope
negative
increasing
positive
none of the above

Question

Accepted Answer

# Brief Explanations:
To determine the position at point $ C $ on a velocity - time graph, we use the concept that the position change is the net area under the velocity - time curve from the start (let's assume $ t = 0 $) to time $ t $ (at point $ C $).

1. First, analyze the areas:
   - From $ t = 0 $ to $ t = 2 $: The velocity is mostly negative (below the time axis) and some positive, but the net area here might be small or negative.
   - From $ t = 2 $ to $ t = 4 $: The velocity is positive (above the time axis), so this area is positive and adds to the position.
   - From $ t = 4 $ to $ t = 6 $: The velocity is negative (below the time axis), so this area is negative and subtracts from the position.
   - From $ t = 6 $ to $ t = 7 $ (around point $ C $): The velocity is just crossing zero (at $ C $, velocity is zero). But overall, when we sum up the areas:
     - The positive area from $ t = 2 $ to $ t = 4 $ and the subsequent areas: Let's think about the net displacement. The area above the time axis (positive velocity) and below (negative velocity). The key is that the net area (displacement) at $ C $:
       - The area from $ t = 0 $ to $ t = 2 $: Let's assume the first part (before $ t = 2 $) has a small negative or near - zero net area. Then from $ t = 2 $ to $ t = 4 $, we have a positive area (since velocity is positive). Then from $ t = 4 $ to $ t = 6 $, we have a negative area (velocity negative). But when we get to $ t = 7 $ (point $ C $), we need to see the cumulative effect. However, another way: The position at a time $ t $ is the integral of velocity from $ 0 $ to $ t $, $ x(t)=\int_{0}^{t}v(t')dt' $.
       - Let's consider the regions:
         - Region 1 ($ t = 0 $ to $ t = 2 $): The velocity curve is below and above the axis, but the net area here is likely small (maybe slightly negative or zero).
         - Region 2 ($ t = 2 $ to $ t = 4 $): Velocity is positive, so area is positive (adds to position).
         - Region 3 ($ t = 4 $ to $ t = 6 $): Velocity is negative, area is negative (subtracts from position).
         - Region 4 ($ t = 6 $ to $ t = 7 $): Velocity is positive (since at $ C $, velocity is zero and before $ C $ (around $ t = 6 $ to $ t = 7 $) the velocity is coming from negative to zero, wait no - looking at the graph, at $ t = 6 $, the velocity is zero (the point $ C $ is on the time axis). Wait, actually, the graph: from $ t = 4 $ to $ t = 6 $, velocity goes negative (down to $ B $) and then comes back up to zero at $ C $ (around $ t = 7 $? Wait the x - axis is $ 0,2,4,6,8,10 $. So $ C $ is at $ t = 7 $ (between 6 and 8) with velocity zero.
         - Now, the net displacement: The area above the axis (positive velocity) and below (negative velocity). The positive area from $ t = 2 $ to $ t = 4 $ and the area from $ t = 6 $ to $ t = 7 $ (positive, since velocity is going from negative to zero, so in $ t = 6 $ to $ t = 7 $, velocity is positive? Wait no, the graph: after $ t = 6 $, the velocity goes from zero (at $ t = 6 $)? Wait no, the point $ C $ is on the time axis (velocity = 0). Wait, maybe I misread. Let's re - analyze:
           - From $ t = 0 $ to $ t = 2 $: The velocity curve starts at $ v=-1 $, goes up, down, and back to zero at $ t = 2 $. So the area here is the integral of $ v(t) $ from 0 to 2. Since part of it is below the axis (negative velocity) and part above, but the net area might be negative or close to zero.
           - From $ t = 2 $ to $ t = 4 $: Velocity is positive (above the axis), so the area is positive (let's say $ A_1>0 $).
           - From $ t = 4 $ to $ t = 6 $: Velocity is negative (below the axis), so the area is negative ( $ A_2<0 $ ).
           - From $ t = 6 $ to $ t = 7 $ (at $ C $): Velocity is zero at $ C $, and just before $ C $ (from $ t = 6 $ to $ t = 7 $), the velocity is positive (since it's coming up from the negative region to zero). Wait, no, the graph: at $ t = 6 $, the velocity is zero (the point $ C $ is on the time axis). Then from $ t = 6 $ to $ t = 8 $, the velocity goes negative again? Wait the graph shows that at $ t = 8 $, velocity is zero. Wait, maybe the key is that the net displacement (position) at $ C $:
             - The sum of the areas: $ \text{Net Area}=\text{Area from }0 - 2+\text{Area from }2 - 4+\text{Area from }4 - 6+\text{Area from }6 - 7 $
             - The area from $ 0 - 2 $: Let's assume it's a small negative (since velocity is negative at the start). The area from $ 2 - 4 $ is positive (large enough). The area from $ 4 - 6 $ is negative (but maybe the positive area from $ 2 - 4 $ is larger than the negative area from $ 4 - 6 $ and the negative area from $ 0 - 2 $). So the net area (displacement) is positive. So the position at $ C $ is positive.

- Now let's analyze the options:
     - "has a horizontal slope": The slope of the velocity - time graph is acceleration. The position at $ C $ is a scalar quantity (displacement), not a slope. So this is wrong.
     - "negative": As we analyzed, the net displacement is positive, so this is wrong.
     - "increasing": The position at a point is a value, not a rate of change. The velocity at $ C $ is zero, but the position itself is a quantity. The term "increasing" is for a function (like position as a function of time, but we are asked about the position at $ C $, not the rate of change of position). So this is wrong.
     - "positive": As per the net area (displacement) analysis, the position at $ C $ is positive.
     - "none of the above": Since "positive" is correct, this is wrong.

# Answer:
D. positive

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