Edit: Adding some subsequent questions to the end of this post:

There is a question that asks, based on the "velocity vs distance already ran" provided, which "distance vs time" (ie velocity) graph best shows the runner's journey during a 100m run?

You can see the applicable graphs here. Note that acceleration is not constant throughout the 100m.

In order to answer this question, instead of eyeballing (the gradient of the curve which was the recommended way to do this), I want to calculate a few points and see which graph has the applicable distance vs time coordinate.

A previous question in the same set of questions that this question came in asked how long it took to run the last 50m. To answer this, I took the average speed in the last 50m by adding the initial and final velocity and dividing by 2 (on the graph, the initial is 12m/s and 11.5m/s, adding gives me 23.5, dividing by 2 gives me 11.75m/s) then time = distance/speed so 50m/11.75m/s = 4.25 seconds.

To answer the current question, I tried to use the same method: for any segment of the run, find average speed by adding initial speed and final speed, then divide by 2; then divide distance of that segment by average speed taken during that segment) for a few points on the velocity vs distance graph in the segment up to the 20m and 12m/s mark because this segment has a linear relationship and has constant acceleration.

I calculated the initial speed in 2 ways and both ways seemed to be wrong. The first way is to always use the starting speed of 0 at rest as the initial speed:

1. At the 5m mark, average speed was (0+3)/2 = 1.5m/s. Time is then 5/1.5 =3.33s

2. At the 10m mark, average speed was (0+6)/2 = 3m/s. Time is then 10/3 =3.33s

3. At the 15m mark, average speed was (0+9)/2 = 4.5m/s. Time is then 15/4.5 =3.33s

This can't be right because he can't have run 5m AND 10m AND 15m in 3.33 seconds (not 3 seconds each because each time I use the same initial velocity) since 15m includes the previous 5m and 10m. But if it were correct, to illustrate what I wanted to do, I would be looking at a graph that has the following coordinates that fall on one of graph lines:

Time of 3.33s and distance of 5m

Time of 3.33s and distance of 10m

Time of 3.33s and distance of 15m

The second method I used was to break the entire section down into individual segments of 5m each:

1. At the first 5m mark, average speed was (0+3)/2 = 1.5m/s. Time is then 5/1.5 =3.33s

2. At the 10m mark which is the second 5m mark or 5m from the previous, average speed was (3+6)/2 = 4.5m/s. Time is then 5/4.5 =1.11s (5 instead of 10 because he ran 5m in this segment).

3. At the 15m mark which is also the third 5m mark, average speed was (6+9)/2 = 7.5m/s. Time is then 5/7.5 =0.66s

4. At the 20m mark which is also the 4th 5m mark, average speed was (9+11)/2 = 10m/s. Time is then 5/10 =0.5s

This is equally impossible because it means the runner would have taken just half a second to run the last 5m when it took him 3.33s to run the first 5m.

If it were correct, I would look for a graph in the answer that has the following coordinates on one of the lines:

Time of 1.5s and distance of 5m

Time of 1.5s (for the previous segment) + 1.11s (for the current segment) = 2.61s and distance of 10m

Time of 2.61s (from previous accumulated time) + 0.66s = 3.27s and distance of 15m.

Time of 3.27 (from previous accumulated time) + 0.5s = 3.77s and distance of 20m.

Graph III has the coordinates (1.5s, 5m), (2.61s, 10m) which are 1 and 2 above respectively, but not for (2.61s, 15m) and (3.77s, 20m) (2 and 4 respectively). Taking (3.77s, 20m) as an example, the 20m mark on graph III is actually less than 3.5m not even close to 3.77 (see images where I drew the green line).

So none of the options suit my calculations.

What am I doing wrong?

(The official answer is actually Graph III).

Can I assume all graphs are to scale, or is it like any other diagram where you cannot usually assume it is to scale and must go by numbers instead? I would have thought graphs by nature would have to be at least in proportion, even if it were stretched along one axis, halfway between 3 and 4 would still be half way, and in graph III the green line is clearly not close to half way, much less three quarters of the way to 4 (3.77).

Extra questions:

1. A later question in this question set asks "The distance that the runner travelled during the first 5.0 seconds is closest to". I know you can find the answer with graph III (if you chose correctly). I would like to know if it is possible to calculate this if you chose wrong or maybe you did not have those graphs.

2. One more question says, "Suppose the runner ran 200m in 19.5 seconds and the first 100m was as shown in the original graph. If his acceleration was constant for the last 100 metres, what was his velocity as he crossed the finishing line?"

The model answer for this used the distance travelled = 1/2*(initial + final velocity)*t approach: specifically, initial velocity of the second 100m is the final velocity of the previous 100m, so 11.5m/s, and we are looking for the final velocity which was done by rearranging the formula. (the answer is 9.55m/s).

This method of finding the average velocity (1/2*(initial + final velocity)) is what I used (and the model answer used) to find the time taken for the last 50m in the first 100m.

Here is my question: We are given the time taken (9.5s) to run a distance (100m), and velocity is distance over time. His acceleration was constant, so why doesn't a simply dividing 100m by 9.5s to find his velocity, work?

There is a question that asks, based on the "velocity vs distance already ran" provided, which "distance vs time" (ie velocity) graph best shows the runner's journey during a 100m run?

You can see the applicable graphs here. Note that acceleration is not constant throughout the 100m.

In order to answer this question, instead of eyeballing (the gradient of the curve which was the recommended way to do this), I want to calculate a few points and see which graph has the applicable distance vs time coordinate.

A previous question in the same set of questions that this question came in asked how long it took to run the last 50m. To answer this, I took the average speed in the last 50m by adding the initial and final velocity and dividing by 2 (on the graph, the initial is 12m/s and 11.5m/s, adding gives me 23.5, dividing by 2 gives me 11.75m/s) then time = distance/speed so 50m/11.75m/s = 4.25 seconds.

To answer the current question, I tried to use the same method: for any segment of the run, find average speed by adding initial speed and final speed, then divide by 2; then divide distance of that segment by average speed taken during that segment) for a few points on the velocity vs distance graph in the segment up to the 20m and 12m/s mark because this segment has a linear relationship and has constant acceleration.

I calculated the initial speed in 2 ways and both ways seemed to be wrong. The first way is to always use the starting speed of 0 at rest as the initial speed:

1. At the 5m mark, average speed was (0+3)/2 = 1.5m/s. Time is then 5/1.5 =3.33s

2. At the 10m mark, average speed was (0+6)/2 = 3m/s. Time is then 10/3 =3.33s

3. At the 15m mark, average speed was (0+9)/2 = 4.5m/s. Time is then 15/4.5 =3.33s

This can't be right because he can't have run 5m AND 10m AND 15m in 3.33 seconds (not 3 seconds each because each time I use the same initial velocity) since 15m includes the previous 5m and 10m. But if it were correct, to illustrate what I wanted to do, I would be looking at a graph that has the following coordinates that fall on one of graph lines:

Time of 3.33s and distance of 5m

Time of 3.33s and distance of 10m

Time of 3.33s and distance of 15m

The second method I used was to break the entire section down into individual segments of 5m each:

1. At the first 5m mark, average speed was (0+3)/2 = 1.5m/s. Time is then 5/1.5 =3.33s

2. At the 10m mark which is the second 5m mark or 5m from the previous, average speed was (3+6)/2 = 4.5m/s. Time is then 5/4.5 =1.11s (5 instead of 10 because he ran 5m in this segment).

3. At the 15m mark which is also the third 5m mark, average speed was (6+9)/2 = 7.5m/s. Time is then 5/7.5 =0.66s

4. At the 20m mark which is also the 4th 5m mark, average speed was (9+11)/2 = 10m/s. Time is then 5/10 =0.5s

This is equally impossible because it means the runner would have taken just half a second to run the last 5m when it took him 3.33s to run the first 5m.

If it were correct, I would look for a graph in the answer that has the following coordinates on one of the lines:

Time of 1.5s and distance of 5m

Time of 1.5s (for the previous segment) + 1.11s (for the current segment) = 2.61s and distance of 10m

Time of 2.61s (from previous accumulated time) + 0.66s = 3.27s and distance of 15m.

Time of 3.27 (from previous accumulated time) + 0.5s = 3.77s and distance of 20m.

Graph III has the coordinates (1.5s, 5m), (2.61s, 10m) which are 1 and 2 above respectively, but not for (2.61s, 15m) and (3.77s, 20m) (2 and 4 respectively). Taking (3.77s, 20m) as an example, the 20m mark on graph III is actually less than 3.5m not even close to 3.77 (see images where I drew the green line).

So none of the options suit my calculations.

What am I doing wrong?

(The official answer is actually Graph III).

Can I assume all graphs are to scale, or is it like any other diagram where you cannot usually assume it is to scale and must go by numbers instead? I would have thought graphs by nature would have to be at least in proportion, even if it were stretched along one axis, halfway between 3 and 4 would still be half way, and in graph III the green line is clearly not close to half way, much less three quarters of the way to 4 (3.77).

Extra questions:

1. A later question in this question set asks "The distance that the runner travelled during the first 5.0 seconds is closest to". I know you can find the answer with graph III (if you chose correctly). I would like to know if it is possible to calculate this if you chose wrong or maybe you did not have those graphs.

2. One more question says, "Suppose the runner ran 200m in 19.5 seconds and the first 100m was as shown in the original graph. If his acceleration was constant for the last 100 metres, what was his velocity as he crossed the finishing line?"

The model answer for this used the distance travelled = 1/2*(initial + final velocity)*t approach: specifically, initial velocity of the second 100m is the final velocity of the previous 100m, so 11.5m/s, and we are looking for the final velocity which was done by rearranging the formula. (the answer is 9.55m/s).

This method of finding the average velocity (1/2*(initial + final velocity)) is what I used (and the model answer used) to find the time taken for the last 50m in the first 100m.

Here is my question: We are given the time taken (9.5s) to run a distance (100m), and velocity is distance over time. His acceleration was constant, so why doesn't a simply dividing 100m by 9.5s to find his velocity, work?

(edited 1 year ago)

Original post by martinthemartian

There is a question that asks, based on the "velocity vs distance already ran" provided, which "distance vs time" (ie velocity) graph best shows the runner's journey during a 100m run?

You can see the applicable graphs here. Note that acceleration is not constant throughout the 100m.

In order to answer this question, instead of eyeballing (the gradient of the curve which was the recommended way to do this), I want to calculate a few points and see which graph has the applicable distance vs time coordinate.

A previous question in the same set of questions that this question came in asked how long it took to run the last 50m. To answer this, I took the average speed in the last 50m by adding the initial and final velocity and dividing by 2 (on the graph, the initial is 12m/s and 11.5m/s, adding gives me 23.5, dividing by 2 gives me 11.75m/s) then time = distance/speed so 50m/11.75m/s = 4.25 seconds.

To answer the current question, I tried to use the same method: for any segment of the run, find average speed by adding initial speed and final speed, then divide by 2; then divide distance of that segment by average speed taken during that segment) for a few points on the velocity vs distance graph in the segment up to the 20m and 12m/s mark because this segment has a linear relationship and has constant acceleration.

I calculated the initial speed in 2 ways and both ways seemed to be wrong. The first way is to always use the starting speed of 0 at rest as the initial speed:

1. At the 5m mark, average speed was (0+3)/2 = 1.5m/s. Time is then 5/1.5 =3.33s

2. At the 10m mark, average speed was (0+6)/2 = 3m/s. Time is then 10/3 =3.33s

3. At the 15m mark, average speed was (0+9)/2 = 4.5m/s. Time is then 15/4.5 =3.33s

This can't be right because he can't have run 5m AND 10m AND 15m in 3.33 seconds (not 3 seconds each because each time I use the same initial velocity) since 15m includes the previous 5m and 10m. But if it were correct, to illustrate what I wanted to do, I would be looking at a graph that has the following coordinates that fall on one of graph lines:

Time of 3.33s and distance of 5m

Time of 3.33s and distance of 10m

Time of 3.33s and distance of 15m

The second method I used was to break the entire section down into individual segments of 5m each:

1. At the first 5m mark, average speed was (0+3)/2 = 1.5m/s. Time is then 5/1.5 =3.33s

2. At the 10m mark which is the second 5m mark or 5m from the previous, average speed was (3+6)/2 = 4.5m/s. Time is then 5/4.5 =1.11s (5 instead of 10 because he ran 5m in this segment).

3. At the 15m mark which is also the third 5m mark, average speed was (6+9)/2 = 7.5m/s. Time is then 5/7.5 =0.66s

4. At the 20m mark which is also the 4th 5m mark, average speed was (9+11)/2 = 10m/s. Time is then 5/10 =0.5s

This is equally impossible because it means the runner would have taken just half a second to run the last 5m when it took him 3.33s to run the first 5m.

If it were correct, I would look for a graph in the answer that has the following coordinates on one of the lines:

Time of 1.5s and distance of 5m

Time of 1.5s (for the previous segment) + 1.11s (for the current segment) = 2.61s and distance of 10m

Time of 2.61s (from previous accumulated time) + 0.66s = 3.27s and distance of 15m.

Time of 3.27 (from previous accumulated time) + 0.5s = 3.77s and distance of 20m.

Graph III has the coordinates (1.5s, 5m), (2.61s, 10m) which are 1 and 2 above respectively, but not for (2.61s, 15m) and (3.77s, 20m) (2 and 4 respectively). Taking (3.77s, 20m) as an example, the 20m mark on graph III is actually less than 3.5m not even close to 3.77 (see images where I drew the green line).

So none of the options suit my calculations.

What am I doing wrong?

(The official answer is actually Graph III).

Can I assume all graphs are to scale, or is it like any other diagram where you cannot usually assume it is to scale and must go by numbers instead? I would have thought graphs by nature would have to be at least in proportion, even if it were stretched along one axis, halfway between 3 and 4 would still be half way, and in graph III the green line is clearly not close to half way, much less three quarters of the way to 4 (3.77).

You can see the applicable graphs here. Note that acceleration is not constant throughout the 100m.

In order to answer this question, instead of eyeballing (the gradient of the curve which was the recommended way to do this), I want to calculate a few points and see which graph has the applicable distance vs time coordinate.

A previous question in the same set of questions that this question came in asked how long it took to run the last 50m. To answer this, I took the average speed in the last 50m by adding the initial and final velocity and dividing by 2 (on the graph, the initial is 12m/s and 11.5m/s, adding gives me 23.5, dividing by 2 gives me 11.75m/s) then time = distance/speed so 50m/11.75m/s = 4.25 seconds.

To answer the current question, I tried to use the same method: for any segment of the run, find average speed by adding initial speed and final speed, then divide by 2; then divide distance of that segment by average speed taken during that segment) for a few points on the velocity vs distance graph in the segment up to the 20m and 12m/s mark because this segment has a linear relationship and has constant acceleration.

I calculated the initial speed in 2 ways and both ways seemed to be wrong. The first way is to always use the starting speed of 0 at rest as the initial speed:

1. At the 5m mark, average speed was (0+3)/2 = 1.5m/s. Time is then 5/1.5 =3.33s

2. At the 10m mark, average speed was (0+6)/2 = 3m/s. Time is then 10/3 =3.33s

3. At the 15m mark, average speed was (0+9)/2 = 4.5m/s. Time is then 15/4.5 =3.33s

This can't be right because he can't have run 5m AND 10m AND 15m in 3.33 seconds (not 3 seconds each because each time I use the same initial velocity) since 15m includes the previous 5m and 10m. But if it were correct, to illustrate what I wanted to do, I would be looking at a graph that has the following coordinates that fall on one of graph lines:

Time of 3.33s and distance of 5m

Time of 3.33s and distance of 10m

Time of 3.33s and distance of 15m

The second method I used was to break the entire section down into individual segments of 5m each:

1. At the first 5m mark, average speed was (0+3)/2 = 1.5m/s. Time is then 5/1.5 =3.33s

2. At the 10m mark which is the second 5m mark or 5m from the previous, average speed was (3+6)/2 = 4.5m/s. Time is then 5/4.5 =1.11s (5 instead of 10 because he ran 5m in this segment).

3. At the 15m mark which is also the third 5m mark, average speed was (6+9)/2 = 7.5m/s. Time is then 5/7.5 =0.66s

4. At the 20m mark which is also the 4th 5m mark, average speed was (9+11)/2 = 10m/s. Time is then 5/10 =0.5s

This is equally impossible because it means the runner would have taken just half a second to run the last 5m when it took him 3.33s to run the first 5m.

If it were correct, I would look for a graph in the answer that has the following coordinates on one of the lines:

Time of 1.5s and distance of 5m

Time of 1.5s (for the previous segment) + 1.11s (for the current segment) = 2.61s and distance of 10m

Time of 2.61s (from previous accumulated time) + 0.66s = 3.27s and distance of 15m.

Time of 3.27 (from previous accumulated time) + 0.5s = 3.77s and distance of 20m.

Graph III has the coordinates (1.5s, 5m), (2.61s, 10m) which are 1 and 2 above respectively, but not for (2.61s, 15m) and (3.77s, 20m) (2 and 4 respectively). Taking (3.77s, 20m) as an example, the 20m mark on graph III is actually less than 3.5m not even close to 3.77 (see images where I drew the green line).

So none of the options suit my calculations.

What am I doing wrong?

(The official answer is actually Graph III).

Can I assume all graphs are to scale, or is it like any other diagram where you cannot usually assume it is to scale and must go by numbers instead? I would have thought graphs by nature would have to be at least in proportion, even if it were stretched along one axis, halfway between 3 and 4 would still be half way, and in graph III the green line is clearly not close to half way, much less three quarters of the way to 4 (3.77).

First question.

You are totally over thinking all this. What you are doing is not the way to tackle this sort of question.

The essence of the first (v-s) graph is that it starts with the runner accelerating.

A distance-time graph of the same motion will show the distance increasing but as time goes on it will increase at a greater rate.

So the line will curve upwards.

Only graphs iii and iv show this.

The v-s graph then shows the velocity levelling out. (It does reduce very slightly.)

So the distance time graph should now show a straight(ish) line with a constant gradient. (The sign of a constant velocity)

Only graph iii shows this.

The logic you employ to solve such a question is based on a knowledge of what a v-t (or v-s) and s-t graph is showing.

You often do this by a process of elimination, as I have just explained. Such non-calculation questions are done by just 'eyeballing'.

I would expect you to be able to answer this purely by inspection in about 4 seconds.

Calculation questions on the other hand, do need you to read values off the graph.

This question was not about calculations.

(edited 1 year ago)

Original post by Stonebridge

First question.

You are totally over thinking all this. What you are doing is not the way to tackle this sort of question.

The essence of the first (v-t) graph is that it starts with the runner accelerating. The straight line shows uniform acceleration.

A distance-time graph of the same motion will show the distance increasing but as time goes on it will increase at a greater rate.

So the line will curve upwards.

Only graphs iii and iv show this.

The v-t graph then shows the velocity levelling out. (It does reduce very slightly.)

So the distance time graph should now show a straight(ish) line with a constant gradient. (The sign of a constant velocity)

Only graph iii shows this.

The logic you employ to solve such a question is based on a knowledge of what a v-t and s-t graph is showing.

You often do this by a process of elimination, as I have just explained. Such non-calculation questions are done by just 'eyeballing'.

I would expect you to be able to answer this purely by inspection in about 4 seconds.

Calculation questions on the other hand, do need you to read values off the graph.

This question was not about calculations.

You are totally over thinking all this. What you are doing is not the way to tackle this sort of question.

The essence of the first (v-t) graph is that it starts with the runner accelerating. The straight line shows uniform acceleration.

A distance-time graph of the same motion will show the distance increasing but as time goes on it will increase at a greater rate.

So the line will curve upwards.

Only graphs iii and iv show this.

The v-t graph then shows the velocity levelling out. (It does reduce very slightly.)

So the distance time graph should now show a straight(ish) line with a constant gradient. (The sign of a constant velocity)

Only graph iii shows this.

The logic you employ to solve such a question is based on a knowledge of what a v-t and s-t graph is showing.

You often do this by a process of elimination, as I have just explained. Such non-calculation questions are done by just 'eyeballing'.

I would expect you to be able to answer this purely by inspection in about 4 seconds.

Calculation questions on the other hand, do need you to read values off the graph.

This question was not about calculations.

Oh I see, thank you Stonebridge.

Can I at least ask why trying to calculate the time taken to run for the first 20m using the same method as when trying to find the time taken to run last 50m didn't work?

Original post by martinthemartian

Oh I see, thank you Stonebridge.

Can I at least ask why trying to calculate the time taken to run for the first 20m using the same method as when trying to find the time taken to run last 50m didn't work?

Can I at least ask why trying to calculate the time taken to run for the first 20m using the same method as when trying to find the time taken to run last 50m didn't work?

The final 50m is pretty easy as we know it is (near enough) a constant velocity.

So you can simply apply the basic s=vt equation, using the average velocity of about 11.75 m/s as you have correctly done, to find t.

For the first 20m things are not so straight forward.

But the information we have from the graph is that

Initial velocity = 0 m/s

Final velocity = 11 m/s

distance = 20 m

We don't know the acceleration from the graph as this is a function of velocity and time, not velocity and distance.

We don't even know if that is a uniform acceleration.

But using distance = average velocity x time for this section gives

20 = (11/2) t

and t= 3.6s

This will be approximate.

Graph 3, which is the given 'correct' graph shows the value to be about 3.5s

The calculation is not straightforward because it's a v-s graph not a v-t graph.

At A-Level in physics you will be given a v-t or s-t graph, not a v-s graph to look at.

There is no 'simple' way to read the graph as the velocity is varying with time and also with distance.

But of course, the distance is also varying with time. The maths is more complicated if the acceleration is not uniform.

At A-Level it will be uniform.

Original post by Stonebridge

The final 50m is pretty easy as we know it is (near enough) a constant velocity.

So you can simply apply the basic s=vt equation, using the average velocity of about 11.75 m/s as you have correctly done, to find t.

For the first 20m things are not so straight forward.

So you can simply apply the basic s=vt equation, using the average velocity of about 11.75 m/s as you have correctly done, to find t.

For the first 20m things are not so straight forward.

Is the final 50m easy only because the acceleration is "almost constant" and the first 20m more difficult because the acceleration is unknown?

Asking because I haven't understood what the "rule" is that is preventing me from using s = vt formula to find time taken (if I wanted to, I realise you are better off just eyeballing the given graphs now) for the first 5, 10 15 and 20m:

the final 50m has an average velocity of 11.75 which can be rounded to 12 which is the same as the initial velocity and therefore could be argued it is "almost constant". But that 0.25m/2 being "constant" can be subjective. How close is close enough? For example, maybe some sort of precise machinery cannot have its velocity deviate more than 0.02m/s so a change of 0.25m/s is a massive change. Visually, if it were a line of best fit, the trend is clearly negative, so when I first saw this question, I wasn't going to say it is constant at all, the fact that the speed in the last 50m was decreasing was as clear to me as the fact that the speed was increasing in the first 20m. It is possible that at a scale large enough, the steepness of the initial 20m goes away and even looks flat and constant: proportionally, if the earth was an apple, the earth's crust is only as thick as the apple's skin and appears flat with a consistent thickness (constant gradient), even though the earth has some very steep and tall mountains.

Also:

using distance = average velocity x time for this section gives

20 = (11/2) t

and t= 3.6s

20 = (11/2) t

and t= 3.6s

Using this method, is the 3.6s the time taken to run the entire first 20m? Is this method "allowed" (considering the first 20m is "not so straight forward"?) If so, can the same method be used for the first 5, 10, and 15m to get the (seemingly unrealistic) answers I already figured out? Is it unrealistic because it is only a mathematical question that is a real-world impossibility or is it actually mathematically wrong?

(edited 1 year ago)

Original post by martinthemartian

Is the final 50m easy only because the acceleration is "almost constant" and the first 20m more difficult because the acceleration is unknown?

Asking because I haven't understood what the "rule" is that is preventing me from using s = vt formula to find time taken (if I wanted to, I realise you are better off just eyeballing the given graphs now) for the first 5, 10 15 and 20m:

the final 50m has an average velocity of 11.75 which can be rounded to 12 which is the same as the initial velocity and therefore could be argued it is "almost constant". But that 0.25m/2 being "constant" can be subjective. How close is close enough? For example, maybe some sort of precise machinery cannot have its velocity deviate more than 0.02m/s so a change of 0.25m/s is a massive change. Visually, if it were a line of best fit, the trend is clearly negative, so when I first saw this question, I wasn't going to say it is constant at all, the fact that the speed in the last 50m was decreasing was as clear to me as the fact that the speed was increasing in the first 20m. It is possible that at a scale large enough, the steepness of the initial 20m goes away and even looks flat and constant: proportionally, if the earth was an apple, the earth's crust is only as thick as the apple's skin and appears flat with a consistent thickness (constant gradient), even though the earth has some very steep and tall mountains.

Also:

Using this method, is the 3.6s the time taken to run the entire first 20m? Is this method "allowed" (considering the first 20m is "not so straight forward"?) If so, can the same method be used for the first 5, 10, and 15m to get the (seemingly unrealistic) answers I already figured out? Is it unrealistic because it is only a mathematical question that is a real-world impossibility or is it actually mathematically wrong?

Asking because I haven't understood what the "rule" is that is preventing me from using s = vt formula to find time taken (if I wanted to, I realise you are better off just eyeballing the given graphs now) for the first 5, 10 15 and 20m:

the final 50m has an average velocity of 11.75 which can be rounded to 12 which is the same as the initial velocity and therefore could be argued it is "almost constant". But that 0.25m/2 being "constant" can be subjective. How close is close enough? For example, maybe some sort of precise machinery cannot have its velocity deviate more than 0.02m/s so a change of 0.25m/s is a massive change. Visually, if it were a line of best fit, the trend is clearly negative, so when I first saw this question, I wasn't going to say it is constant at all, the fact that the speed in the last 50m was decreasing was as clear to me as the fact that the speed was increasing in the first 20m. It is possible that at a scale large enough, the steepness of the initial 20m goes away and even looks flat and constant: proportionally, if the earth was an apple, the earth's crust is only as thick as the apple's skin and appears flat with a consistent thickness (constant gradient), even though the earth has some very steep and tall mountains.

Also:

Using this method, is the 3.6s the time taken to run the entire first 20m? Is this method "allowed" (considering the first 20m is "not so straight forward"?) If so, can the same method be used for the first 5, 10, and 15m to get the (seemingly unrealistic) answers I already figured out? Is it unrealistic because it is only a mathematical question that is a real-world impossibility or is it actually mathematically wrong?

The problem I'm having with all these questions is knowing exactly what you have learned and the context of these questions.

My answers will depend on that.

I have no idea what is 'acceptable' as I have no idea at what level you are doing this, or what course this is.

If the original question was simply to say which of the 4 graphs was the correct answer, I think we have covered that.

The equations studied at A-level are for constant (or uniform) acceleration. As well as the simpler case of constant velocity.

For constant acceleration you can use: distance = average velocity x time

this is a standard formula and is simply an extension of the formula s=vt, substituting average velocity for velocity. This is mathematically correct.

Have you seen this proved graphically?

In the question, in the given v-s graph, over that last 50m the velocity is hardly changing.

It would be quite acceptable to use distance = average velocity x time there to find the time taken.

In the first section of the v-s graph, the first 20m, things are more complicated.

You can't use the standard so called 'suvat' equations for uniform acceleration. If it was a v-t graph it would be straightforward.

Have you learned these and applied them to various examples of uniformly accelerated motion?

What I've done is to make an approximation. Whether or not that is 'acceptable' I don't know without further information on the context and level of the question.

Without some context on where this question has come from and to what level the calculations are expected to be done (if at all!) it is almost impossible to give you an exact answer.

Original post by Stonebridge

The problem I'm having with all these questions is knowing exactly what you have learned and the context of these questions.

My answers will depend on that.

I have no idea what is 'acceptable' as I have no idea at what level you are doing this, or what course this is.

If the original question was simply to say which of the 4 graphs was the correct answer, I think we have covered that.

The equations studied at A-level are for constant (or uniform) acceleration. As well as the simpler case of constant velocity.

For constant acceleration you can use: distance = average velocity x time

this is a standard formula and is simply an extension of the formula s=vt, substituting average velocity for velocity. This is mathematically correct.

Have you seen this proved graphically?

In the question, in the given v-s graph, over that last 50m the velocity is hardly changing.

It would be quite acceptable to use distance = average velocity x time there to find the time taken.

In the first section of the v-s graph, the first 20m, things are more complicated.

You can't use the standard so called 'suvat' equations for uniform acceleration. If it was a v-t graph it would be straightforward.

Have you learned these and applied them to various examples of uniformly accelerated motion?

What I've done is to make an approximation. Whether or not that is 'acceptable' I don't know without further information on the context and level of the question.

Without some context on where this question has come from and to what level the calculations are expected to be done (if at all!) it is almost impossible to give you an exact answer.

My answers will depend on that.

I have no idea what is 'acceptable' as I have no idea at what level you are doing this, or what course this is.

If the original question was simply to say which of the 4 graphs was the correct answer, I think we have covered that.

The equations studied at A-level are for constant (or uniform) acceleration. As well as the simpler case of constant velocity.

For constant acceleration you can use: distance = average velocity x time

this is a standard formula and is simply an extension of the formula s=vt, substituting average velocity for velocity. This is mathematically correct.

Have you seen this proved graphically?

In the question, in the given v-s graph, over that last 50m the velocity is hardly changing.

It would be quite acceptable to use distance = average velocity x time there to find the time taken.

In the first section of the v-s graph, the first 20m, things are more complicated.

You can't use the standard so called 'suvat' equations for uniform acceleration. If it was a v-t graph it would be straightforward.

Have you learned these and applied them to various examples of uniformly accelerated motion?

What I've done is to make an approximation. Whether or not that is 'acceptable' I don't know without further information on the context and level of the question.

Without some context on where this question has come from and to what level the calculations are expected to be done (if at all!) it is almost impossible to give you an exact answer.

I am sorry I missed this reply.

I did not study physics at the secondary level, and these questions come "as is" and I have already posted all of the relevant information for the questions, there are no more "before" information for further context (each set of question is independent of the previous and future sets and can be from any topic). They are part of the GAMSAT practice booklets (GAMSAT is a graduate entry exam for studying medicine) and for the purposes of the exam, I learnt physics concepts from various sources - text books, videos, websites, asking questions. The thing about the questions in GAMSAT booklets is that they are not your typical academic questions where you are asked to do calculations, the wording is often very bad, the questions themselves ask for things you would not calculate or do in real life and as you have seen, the answers are often estimates of the real answer (eg 4.3 instead of 4.25) because you are not allowed to use calculators in the exam.

I have studied maths up to first year university, as such, I have no problem doing the actual mathematical calculations once I figure out what needs to be done, but sometimes I do not know what needs to be done from a physics concept point of view.

In the case of my last question, for example, I calculated the average velocity for the last 50m by adding together the initial and final velocity and dividing by 2 = and using that to find the time taken. I used the same method of calculation for the first 20m but it didn't seem to work, so it seems there is a hole in my understanding here and I am not sure what it is. While I agree that the last 50m is almost constant, I would also argue that "almost" can be subjective. For example, at what steepness would the gradient be considered close to being constant, and what gradient would it be considered no longer constant? To me it was all the same: the first 20m in the v-d graph is a straight line with no curves just like the last 50m, so something (perhaps acceleration?) must have been constant, so I thought (v+u)/2 would be a quick and easy way to get the average velocity.

I have learnt SUVAT and so far all the straight forward practice questions I have come across, I can do using the SUVAT formulae.

For constant acceleration you can use: distance = average velocity x time

this is a standard formula and is simply an extension of the formula s=vt, substituting average velocity for velocity. This is mathematically correct.

Have you seen this proved graphically?

this is a standard formula and is simply an extension of the formula s=vt, substituting average velocity for velocity. This is mathematically correct.

Have you seen this proved graphically?

I don't think I have seen this proven graphically. If you have a link to a video where I can see this, it would be much appreciated.

Thanks again.

Original post by martinthemartian

........

I don't think I have seen this proven graphically. If you have a link to a video where I can see this, it would be much appreciated.

Thanks again.

I don't think I have seen this proven graphically. If you have a link to a video where I can see this, it would be much appreciated.

Thanks again.

Actually, you should be able to derive since you have studied maths up to university.

The maths is mainly gcse level.

We can find the displacement or distance travelled as area under the graph from the diagram above which is

$s = \dfrac{1}{2} (u + v)t$

Note that average speed is total distance/total time which is

$v_{\text{ave}} = \dfrac{1}{2} (u + v)$

For constant acceleration,

Total displacement is the product average velocity and total time taken.

Speed tells us how fast something or someone is traveling. You can find the average rate of an object if you know the distance traveled and the time it took.

The formula for speed is speed = distance ÷ time. To determine the units for speed, you need to know the units for distance and time. In this example, distance is in meters (m) and time is in seconds (s), so the units will be in meters per second (m/s).

Time based on distance and velocity is speed = distance ÷ time. distance = speed × time. time = distance ÷ speed.

The formula for speed is speed = distance ÷ time. To determine the units for speed, you need to know the units for distance and time. In this example, distance is in meters (m) and time is in seconds (s), so the units will be in meters per second (m/s).

Time based on distance and velocity is speed = distance ÷ time. distance = speed × time. time = distance ÷ speed.

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