Need help with the interpretation of this v(t) graph

In summary,-Determine the distance traveled by the object between 0-7 s.-Determine the acceleration of the piece in the range of 0-4 s-Determine the acceleration of the piece in the range of 4-7 s.
  • #1
gladidi
19
4
Summary:: Please tell me an example

Determine the distance traveled by the object between 0-7 s.

Determine the acceleration of the piece in the range of 0-4 s
Determine the acceleration of the piece in the range of 4-7 s.I've tried, but something I'm doing wrong. Could you clarify?
hhh.PNG
 
Physics news on Phys.org
  • #2
Welcome to PF.

You need to show us your work before we can help you with this. How is the velocity as a function of time related to the position and acceleration as a function of time? How can you use that to help you solve these questions?
 
  • #3
berkeman said:
Welcome to PF.

You need to show us your work before we can help you with this. How is the velocity as a function of time related to the position and acceleration as a function of time? How can you use that to help you solve these questions?
This is the work.
 

Attachments

  • Screenshot_20210910_221318.jpg
    Screenshot_20210910_221318.jpg
    33.8 KB · Views: 86
  • #4
gladidi said:
This is the work.
No, that is the graph you are supposed to use to figure out those 3 questions.

Please answer this:
berkeman said:
How is the velocity as a function of time related to the position and acceleration as a function of time?
 
  • #5
berkeman said:
No, that is the graph you are supposed to use to figure out those 3 questions.

Please answer this:
I've tried to solve it with a chart, but the answers aren't going right
 
  • #6
You didn't answer my questions. Until you do that, you will not get very far on problems like this one...
 
  • Like
Likes Chestermiller
  • #7
How is velocity defined in terms of position and time?

How is acceleration defined in terms of velocity and time?
 
  • #8
gladidi said:
Summary::
I've tried, but something I'm doing wrong. Could you clarify?
View attachment 288876
What specifically did you try?

gladidi said:
I've tried to solve it with a chart, but the answers aren't going right
What approach with a chart did you try?
What were your answers
and how did you get them, specifically?
 
  • #9
robphy said:
What specifically did you try?What approach with a chart did you try?
What were your answers
and how did you get them, specifically?
> Determine the acceleration of the piece in the range of 0-4 s

< 12/4
 
  • #10
gladidi said:
> Determine the acceleration of the piece in the range of 0-4 s

< 12/4
Ok, that's a start.
Why 12/4?
 
  • Like
Likes berkeman
  • #11
robphy said:
Ok, that's a start.
Why 12/4?
Speed change / time to change
 
  • #12
robphy said:
Ok, that's a start.
Why 12/4?
Acceleration formula
 
  • #13
gladidi said:
Speed change / time to change

That sounds like a slope.
What is the speed-change [although velocity-change is more correct], during the time-change?
 
  • #14
gladidi said:
Speed change / time to change
What is the speed at t=0s?
What is the speed at t=4s?
What is the change in speed?
 
  • #15
haruspex said:
What is the speed at t=0s?
What is the speed at t=4s?
What is the change in speed?
Speed at t=0s =0?
Speed at t=4s? =12?
 
  • #16
gladidi said:
Speed at t=0s =0?
Not according to the graph.
 
  • #17
haruspex said:
Not according to the graph.
It is 3?
 
  • #18
gladidi said:
It is 3?
Yes.
 
  • #19
haruspex said:
Yes.
finally i got this right..
 
  • #20
haruspex said:
What is the speed at t=0s?
What is the speed at t=4s?
What is the change in speed?
How solve this? Determine the acceleration of the piece in the range of 4-7 s.
 
  • #21
gladidi said:
How solve this? Determine the acceleration of the piece in the range of 4-7 s.
Same process. What was the increase in velocity in that interval?
 
  • #22
haruspex said:
Same process. What was the increase in velocity in that interval?
4 and 7 have same speed 12?
 
  • #23
gladidi said:
4 and 7 have same speed 12?
Yes, so what was the increase in speed?
 
  • #24
haruspex said:
Yes, so what was the increase in speed?
24?
 
  • #25
gladidi said:
24?
If you are earning $20 an hour and that changes to $20 an hour, how much is your pay rise?
 
  • #26
I give up
 
  • Haha
Likes Steve4Physics, berkeman and Delta2
  • #27
haruspex said:
If you are earning $20 an hour and that changes to $20 an hour, how much is your pay rise?
0...
 
  • Like
Likes Delta2
  • #28
gladidi said:
0...
Right.
So what is the speed increase from t=4s to t=7s?
 
  • #29
haruspex said:
Right.
So what is the speed increase from t=4s to t=7s?
Answer is 0..
 
  • Like
Likes Steve4Physics, TSny and Delta2
  • #30
The last point is to calculate the distance traveled by objects.
it is said that should calculate the surface area
 
  • #31
gladidi said:
The last point is to calculate the distance traveled by objects.
it is said that should calculate the surface area
Yes, in a velocity-time graph the distance traveled is given by the 'area' under the graph line. But the area of each square means the product of the coordinate intervals.
Here, the vertical grid lines are at 1 second intervals and the horizontal grid lines at intervals of 2m/s, so each little square represents 1 second x 2m/s = 2m.
 
  • #32
haruspex said:
Yes, in a velocity-time graph the distance traveled is given by the 'area' under the graph line. But the area of each square means the product of the coordinate intervals.
Here, the vertical grid lines are at 1 second intervals and the horizontal grid lines at intervals of 2m/s, so each little square represents 1 second x 2m/s = 2m.
Is this right way to solve this? example 35*2+8/2*2?
 

Attachments

  • ....PNG
    ....PNG
    13.3 KB · Views: 99
  • #33
gladidi said:
Is this right way to solve this? example 35*2+8/2*2?
That is the right answer. I cannot tell whether you are expected to do it by counting squares are algebraically. I would use standard formulae for areas of rectangles and triangles.
 
  • #34
haruspex said:
That is the right answer. I cannot tell whether you are expected to do it by counting squares are algebraically. I would use standard formulae for areas of rectangles and triangles.
35*2+8/2*2? is 78m but it's not right..
 
  • #35
gladidi said:
35*2+8/2*2? is 78m but it's not right..
You want the distance for the whole 8 seconds, right?
In the first 4 seconds the average speed is (3+12)/2=7.5m/s. Over four seconds that covers 30m.
For the remaining 4 seconds the speed is 12m/s, giving a distance of 48m.
30m+48m=78m.
 

FAQ: Need help with the interpretation of this v(t) graph

What is v(t) graph?

A v(t) graph is a graph that shows the relationship between velocity (v) and time (t). It is a visual representation of how an object's velocity changes over a specific period of time.

How do I interpret a v(t) graph?

To interpret a v(t) graph, you need to look at the slope of the line. A steeper slope indicates a higher velocity, while a flatter slope indicates a lower velocity. You can also determine the direction of the velocity by looking at the sign of the slope (positive for moving forward, negative for moving backward).

What does the x-axis and y-axis represent on a v(t) graph?

The x-axis represents time (t), and the y-axis represents velocity (v). The graph shows how the velocity changes over time.

How can I use a v(t) graph to calculate acceleration?

Acceleration can be calculated by finding the change in velocity (Δv) over the change in time (Δt). This can be done by finding the slope of the v(t) graph.

What are the units of measurement for velocity and time on a v(t) graph?

The units of measurement for velocity are typically meters per second (m/s), and the units for time are usually seconds (s). However, this may vary depending on the specific graph and the units used for the data.

Similar threads

Back
Top