Understanding Distance and Displacement on Velocity-Time Graphs

[rudransh verma]

Homework Statement

From the graph evaluate-
1) average velocity for first 6sec
2) average speed for first 6sec
3) average acceleration from1 to 4sec
4)instantaneous a at 3rd sec
5) plot a-t and x-t graph assuming x=0,t=0

Relevant Equations

${v_{avg}}=displacement/time$
$average speed= distance/time$
Etc…

3) and 4) is easy. Average and instantaneous acceleration is same from 1 to 4 sec since it’s constant acceleration.

1) and 2) I am unable to get the correct area under the curve.
$s = \frac 12*(-5)*1 + \frac12*10*2 + 10*1 + \frac12*10*1$
Same I guess will be distance.

5) what is the graph of x-t
I have attached the graph below.

 

[Orodruin]

Homework Statement:: From the graph evaluate-
1) average velocity for first 6sec
2) average speed for first 6sec
3) average acceleration from1 to 4sec
4)instantaneous a at 3rd sec
5) plot a-t and x-t graph assuming x=0,t=0
Relevant Equations:: ${v_{avg}}=displacement/time$
$average speed= distance/time$
Etc…

1) and 2) I am unable to get the correct area under the curve.

Please show us what you did

 

[bob012345]

Look very carefully at the first part of your calculation of the area under the curve.

 

[Steve4Physics]

Homework Statement:: From the graph evaluate ...
.I have attached the graph below.

The graph has some errors. It shows the initial velocity to be about -6m/s, so:
from t=0 to t=2s, Δv = 6m/s
from t=2s to t=4s Δv = 10m/s

These values won’t give a straight graph from t=0 to t=4s. But the graph shows a straight line (implying constant acceleration).

Also (but less important), values marked on the time-axis are incorrectly spaced.

 

[bob012345]

The graph has some errors. It shows the initial velocity to be about -6m/s, so:
from t=0 to t=2s, Δv = 6m/s
from t=2s to t=4s Δv = 10m/s

These values won’t give a straight graph from t=0 to t=4s. But the graph shows a straight line (implying constant acceleration).

Also (but less important), values marked on the time-axis are incorrectly spaced.

The graph is graphically misleading but that may be a part of the problem to correctly find the $y$ intercept which the OP did in the initial calculation.

 

[Steve4Physics]

The graph is graphically misleading but that may be a part of the problem to correctly find the $y$ intercept which the OP did in the initial calculation.

The OP appears to have used v=-5m/s when t=0. Unless I’m missing something, I don’t see how this value could be obtained calculated.

The (bad) graph shows the initial velocity to be about -6m/s. Maybe the OP has inaccurately read this as -5m/s

It would be interesting to hear from the OP where -5m/s comes from.

 

[bob012345]

The OP appears to have used v=-5m/s when t=0. Unless I’m missing something, I don’t see how this value could be obtained calculated.

The (bad) graph shows the initial velocity to be about -6m/s. Maybe the OP has inaccurately read this as -5m/s

It would be interesting to hear from the OP where -5m/s comes from.

Now, I could have sworn it said $s= \frac {1}{2} (-10) * 1$ originally and my hint was to look at that term specifically and notice that $t≠1$. Assuming it is a straight line the intercept is $-10$ because you can get the slope from the upper part of the line. I think it was edited.

 

[Steve4Physics]

Now, I could have sworn it said $s= \frac {1}{2} (-10) * 1$ originally and my hint was to look at that term specifically and notice that $t≠1$. Assuming it is a straight line the intercept is $-10$ because you can get the slope from the upper part of the line. I think the OP edited it because is says Last edited: Today, 10:34 AM.

To further complicate matters, note that (in Post #1) the OP says:
“Average and instantaneous acceleration is same from 1 to 4 sec “

This suggests thar the graph is meant to be straight from 1 to 4seconds rather than from 0 to 4 seconds. Maybe the t=2s mark on the graph's time-axis is meant to be t=1s?

It really needs the OP to clear-up these issues.

 

[bob012345]

To further complicate matters, note that (in Post #1) the OP says:
“Average and instantaneous acceleration is same from 1 to 4 sec “

This suggests thar the graph is meant to be straight from 1 to 4seconds rather than from 0 to 4 seconds. Maybe the t=2s mark on the graph's time-axis is meant to be t=1s?

It really needs the OP to clear-up these issues.

I think the graph is drawn that way on purpose so the student would have to do the math to get the answers and not make assumptions.

 

[Steve4Physics]

I think the graph is drawn that way on purpose so the student would have to do the math to get the answers and not make assumptions.

It's possible. But the student must then make a different assumption - that the intercept is not at -6m/s even though this can be read-off the graph. I.e. they must assume the graph is deliberately drawn inaccurately.

I'm more inclined to believe it's a mistake. Maybe @rudransh verma can clarify.

 

[rudransh verma]

This suggests thar the graph is meant to be straight from 1 to 4seconds rather than from 0 to 4 seconds. Maybe the t=2s mark on the graph's time-axis is meant to be t=1s?

Yes it looks that the graph is drawn incorrectly. time intervals are unevenly spaced. It should be -6 instead of -5. But what can I do ? This is the question.

 

[Steve4Physics]

Yes it looks that the graph is drawn incorrectly. time intervals are unevenly spaced. It should be -6 instead of -5. But what can I do ?

Alternatives that the spring to mind (in order of preference) are:

a) Point out the problem(s) to you teacher and ask how to proceed.

b) Don’t answer but provide your reasons for not answering.

c) Make any required assumption(s), e.g. that the graph should start at (0, -10), or that the graph starts at (0, -6) so is not straight in the first 6s. State your chosen assumption(s) and answer the questions based on these assumptions.

 

[rudransh verma]

Alternatives that the spring to mind (in order of preference) are:

a) Point out the problem(s) to you teacher and ask how to proceed.

b) Don’t answer but provide your reasons for not answering.

c) Make any required assumption(s), e.g. that the graph should start at (0, -10), or that the graph starts at (0, -6) so is not straight in the first 6s. State your chosen assumption(s) and answer the questions based on these assumptions.

Maybe you can figure it out by looking at the answer given.
For $v_{avg}$ , $[-10*2*\frac12+(4+1)*\frac12*10]*\frac16$.
For $s_{avg}$, $[10*2*\frac12+ (4+1)*\frac12*10]\frac16$

 

[Steve4Physics]

Maybe you can figure it out by looking at the answer given.
For $v_{avg}$ , $[-10*2*\frac12+(4+1)*\frac12*10]*\frac16$.
For $s_{avg}$, $[10*2*\frac12+ (4+1)*\frac12*10]\frac16$

Yes I can. The question is can you work out what the graph should be from the official solution?

Hint: If the official solution puzzles you, note that it uses the formula for the area of a trapezium for the graph between t=2s to t=6s. But, of course, considering this area as 2 triangles and a rectangle will give exactly the same answer.

 

[rudransh verma]

The question is can you work out what the graph should be from the official solution?

 

[Steve4Physics]

Yes. Well done!

 

[rudransh verma]

Yes. Well done!

But
For ${v_{avg}}$ , $[−10∗2∗\frac12…$
For ${s_{avg}}$ , $[10*2*\frac12…$
So is it like the graph above x-axis gives distance in +ve direction and below gives distance in -ve direction. Because for displacement they have subtracted in given answer and for distance they have added up.

 

[Steve4Physics]

But
For ${v_{avg}}$ , $[−10∗2∗\frac12…$
For ${s_{avg}}$ , $[10*2*\frac12…$
So is it like the graph above x-axis gives distance in +ve direction and below gives distance in -ve direction. Because for displacement they have subtracted in given answer and for distance they have added up.

Yes. Displacements, velocities and accelerations are vectors - they have directions which must be taken into accounr. Distance and speeds are scalars so they have no directions.

For example you walk 10m left and then 10m right (so you are back at your start-point), and it takes 20s total.

Complete the blanks:
Total distance = ________ and average speed = _______
Total displacement = ________ and average velocity = _______
Can you complete these?

 

[rudransh verma]

@Steve4Physics But the integration of velocity with respect to time should give displacement not distance.

 

[Steve4Physics]

@Steve4Physics But the integration of velocity with respect to time should give displacement not distance.

Draw the speed-time graph for the original question. What is the key difference between it and the velocity-time graph?

You haven't answered my Post #18 questions, so I am still not sure that you undestand the difference between displacement and distance.

 

[rudransh verma]

Draw the speed-time graph for the original question. What is the key difference between it and the velocity-time graph?

You haven't answered my Post #18 questions, so I am still not sure that you undestand the difference between displacement and distance.

Displacement is the length between initial and final point of the path. Distance is the length of the path.

 

[Steve4Physics]

Displacement is the length between initial and final point of the path. Distance is the length of the path.

Yes. And the speed-time graph is correct. But you still haven't answered my Post #18 questions.

 

[rudransh verma]

Complete the blanks:
Total distance = ________ and average speed = _______
Total displacement = ________ and average velocity = _______
Can you complete these?

20,1
0,0

 

[Steve4Physics]

20,1
0,0

Well done. Only 2 marks lost (do you know why?)

In the original (Post #1) question, during the first 2 seconds the displacement is negative. It is:(-10m/s) * (2s) * ½ = -10m.

The object moved 10m ‘backwards’ (while slowing down).

When finding the total displacement, the part for the first 2s is negative.

When finding the total distance, the part for the first 2s is positive.

average velocity= (total displacement)/(total time)
= [-10*2* ½ + …] / 6

average speed= (total distance)/(total time)
= [10*2* ½ + …] / 6

 

[rudransh verma]

Well done. Only 2 marks lost (do you know why?)

In the original (Post #1) question, during the first 2 seconds the displacement is negative. It is:(-10m/s) * (2s) * ½ = -10m.

The object moved 10m ‘backwards’ (while slowing down).

When finding the total displacement, the part for the first 2s is negative.

When finding the total distance, the part for the first 2s is positive.

average velocity= (total displacement)/(total time)
= [-10*2* ½ + …] / 6

average speed= (total distance)/(total time)
= [10*2* ½ + …] / 6

So using the speed time graph I can only find distance but using the velocity time graph I can find the distance and displacement because there is information about both the parts +movements and -ve movements. Here v-t graph is divided into two parts one is -displacement or distance from 0-2 and the other is +displacement from 2-6.

 

[Steve4Physics]

So using the speed time graph I can only find distance but using the velocity time graph I can find the distance and displacement because there is information about both the parts +movements and -ve movements. Here v-t graph is divided into two parts one is -displacement or distance from 0-2 and the other is +displacement from 2-6.

Yes. That's exactly right.

By the way, you lost 2 marks (in your Post #23 answer) for forgetting units. (20m, 1m/s)