Negative acceleration and instantaneous velocity on an object.

So you need to find the slope of the line at t=45s, which would be the instantaneous velocity at that point.
  • #1
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Homework Statement



The position of an object along a straight tunnel as a function of time is plotted below:

graphxt5.jpg


1. Are there any time intervals in which the object has a negative acceleration? If so, list them and explain how you know.

2. Calculate the instantaneous velocity at t=20s and t-45s.

Homework Equations



N/A

The Attempt at a Solution



So, for the first question I stated that at 30s to 37s the object's velocity is decreasing, thus the acceleration is negative. However, I was told that this is only part of the interval. What am I missing?

For the second question I did the following calculations:

at 20s = (6 m)/(20s) = 0.3 m/s
at 45s = (16m)/(45s) = 0.36 m/s

I did get the instantaneous velocity at 20s correct but not the instantaneous velocity at 45 s. Where am I going wrong?

Thanks for your assistance.
 
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  • #2
I would say that the velocity is decreasing from about 27 seconds (at the inflection point) to about 46s (at the other inflection point). Draw tangent lines to the graph, and note the parts where those slopes are getting smaller and smaller (the slope of the slopes).

I think you must be close for the one at 45 seconds. It should be around what you got, I'd say more around .43m/s though it's an estimate when you are using a graph rather than the function.
 
  • #3
CaptFormal said:

Homework Statement


So, for the first question I stated that at 30s to 37s the object's velocity is decreasing, thus the acceleration is negative. However, I was told that this is only part of the interval. What am I missing?

The velocity is decreasing up until 45 s. Remember that velocity can be negative, so if velocity goes from -1 m/s to -2 m/s, it's decreasing.
For the second question I did the following calculations:

at 20s = (6 m)/(20s) = 0.3 m/s
at 45s = (16m)/(45s) = 0.36 m/s

I did get the instantaneous velocity at 20s correct but not the instantaneous velocity at 45 s. Where am I going wrong?

The velocity should be negative, not positive. Also, you're calculating average velocity. The question is looking for instantaneous velocity, which is the slope of the tangent of the graph.
 

1. What is negative acceleration?

Negative acceleration, also known as deceleration, is when the velocity of an object decreases over time. This can be caused by a force acting in the opposite direction of the object's motion.

2. How is negative acceleration calculated?

Negative acceleration can be calculated by dividing the change in velocity by the change in time. This is represented by the formula a = (vf - vi) / t, where "a" is acceleration, "vf" is final velocity, "vi" is initial velocity, and "t" is time.

3. Can an object have negative acceleration and positive velocity?

Yes, an object can have negative acceleration and positive velocity. This would mean that the object is slowing down, but still moving in a positive direction.

4. What is instantaneous velocity?

Instantaneous velocity is the velocity of an object at a specific moment in time. It is calculated by finding the slope of the tangent line on a position vs. time graph at a specific point.

5. How are negative acceleration and instantaneous velocity related?

Negative acceleration can affect the instantaneous velocity of an object. If an object is experiencing negative acceleration, its instantaneous velocity will decrease over time. This can be seen on a velocity vs. time graph as a downward sloping line.

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