# Understanding Distance and Displacement on Velocity-Time Graphs

• rudransh verma
In summary: 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.In summary, the OP claims that average and instantaneous acceleration is the same from 1 to 4 seconds, but the graph is misleading and the intercept is -10 instead of -6m/s.
rudransh verma
Gold Member
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.

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rudransh verma said:
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

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

rudransh verma said:
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.

Steve4Physics said:
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.

bob012345 said:
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.

SammyS
Steve4Physics said:
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.

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bob012345 said:
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.

Steve4Physics said:
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.

bob012345 said:
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.

Steve4Physics said:
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.

rudransh verma said:
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.

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.

Steve4Physics said:
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.

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##

rudransh verma said:
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.

Steve4Physics said:
The question is can you work out what the graph should be from the official solution?

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Yes. Well done!

Steve4Physics said:
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.

rudransh verma said:
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?

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

rudransh verma said:
@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.

Steve4Physics said:
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.

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rudransh verma said:
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
Steve4Physics said:
Complete the blanks:
Total distance = ________ and average speed = _______
Total displacement = ________ and average velocity = _______
Can you complete these?
20,1
0,0

rudransh verma said:
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

Steve4Physics said:
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.

rudransh verma said:
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)

rudransh verma

## 1. What is the difference between distance and displacement?

Distance is the total length of the path traveled by an object, while displacement is the straight-line distance between the starting and ending points of an object's motion.

## 2. How are distance and displacement represented on a velocity-time graph?

Distance is represented by the total area under the curve on a velocity-time graph, while displacement is represented by the vertical distance between the starting and ending points on the graph.

## 3. Can an object have a positive displacement and a negative distance on a velocity-time graph?

Yes, an object can have positive displacement if it moves in a positive direction (e.g. right) and negative distance if it moves in a negative direction (e.g. left) on a velocity-time graph.

## 4. How can you calculate the average velocity from a velocity-time graph?

The average velocity can be calculated by finding the slope of the line connecting two points on the velocity-time graph. The slope is equal to the change in displacement divided by the change in time.

## 5. What does a horizontal line on a velocity-time graph represent?

A horizontal line on a velocity-time graph represents a constant velocity, where the object is neither accelerating nor decelerating. The slope of a horizontal line is equal to zero, indicating that there is no change in velocity over time.

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