EmRep13
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I have a question where I have s,v and a. I need to find t. I used s = vt - 1/2at^2 and got two answers. The book used v^2 = u" +2as to find u then used v= u + at to find t, so got one answer, which I also got as one of my two. My question is - are there certain times you use certain equations? Like obviously, if it said what are the possible value(s) you'd use the one that gives a quadratic? If not - then why did I get two answers? The situation in question is about a car travelling from A to C in a certain time, accelerating.

Edit: Grammar
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Joseph Green
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Can you send a picture of the question? Just curious about it
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EmRep13
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(Original post by Joseph Green)
Can you send a picture of the question? Just curious about it
Yeah - I was going to type it out for context but felt a bit lazy . Guess I have to now!

A car is travelling along a straight horizontal road with constant acceleration. The car passes over three consecutive points A, B and C where AB = 100m and BC = 300m. The speed of the car at B is 14m/s and the speed of the car at C is 20m/s. Find:

a) The acceleration of the car
b) the time taken for the car to travel from A to C

The answers are 0.34ms^-2 and 25.5 s (3sf) respectively. Typed out because of aforementioned laziness. Thanks!
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Joseph Green
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So from double checking what I noticed is that using the vt - 1/2at^2 typically leads to these kinds of situations where you get 2 possible answers. The reason is the minus sign. A good way to think about it using your moving car is to switch it to a projectile being thrown up into the air. The acceleration is negative and it ends up passing through a chosen height twice (except for the apex of the path). Although it seems silly, that's what is essentially being drawn when you use the vt - 1/2at^2 equation. It does happen if you use the ut + 1/2at^2 where your acceleration is negative originally but it is easier to perceive that way. In these cases where you get 2 possible answers the idea is to choose the more reasonable answer. The reason the textbook used their method was it made it faster to calculate the overall time as opposed to having to calculate 2 different times and summing them together. I hope this helps. If not feel free to dm me and I'll try draw some examples
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EmRep13
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(Original post by Joseph Green)
So from double checking what I noticed is that using the vt - 1/2at^2 typically leads to these kinds of situations where you get 2 possible answers. The reason is the minus sign. A good way to think about it using your moving car is to switch it to a projectile being thrown up into the air. The acceleration is negative and it ends up passing through a chosen height twice (except for the apex of the path). Although it seems silly, that's what is essentially being drawn when you use the vt - 1/2at^2 equation. It does happen if you use the ut + 1/2at^2 where your acceleration is negative originally but it is easier to perceive that way. In these cases where you get 2 possible answers the idea is to choose the more reasonable answer. The reason the textbook used their method was it made it faster to calculate the overall time as opposed to having to calculate 2 different times and summing them together. I hope this helps. If not feel free to dm me and I'll try draw some examples
I thought that but since the acceleration is positive, surely the car doesn't double back? Also your last sentence - "summing them together" - what do you mean?
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Zoeva123
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When you’re choosing which equation to use you should list out what the question has already given you.
Like:

S=
U=
V=
A=
T=

Your question will always give 3 of the above, the other one is the one you’re calculating and the last one is irrelevant.

So for part a) you’re given s,u and v. You need to calculate a. The irrelevant one is t so you find an equation that does not involve t. The only one is v^2= u^2 +2as which is the accurate formula to be used.

For part b) you have all of the components so you can really use any equation to calculate the time but since there is a possibility of getting part a) incorrect you should choose an equation that doesn’t involve acceleration. ( you said the book used v=u+at but personally I would use s=1/2(u+v)t

This method helps answer any suvat question for a level and gcse maths.
Hope this helps and if you’ve got any other questions let me know x
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Joseph Green
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(Original post by EmRep13)
I thought that but since the acceleration is positive, surely the car doesn't double back? Also your last sentence - "summing them together" - what do you mean?
Summing together is just a fancy way of saying adding. And like I said it's really odd to think about but remember the minus sign makes all the difference here when it comes to drawing the parabola and extrapolating results from it. It's true that the acceleration is positive and it doesn't double back on the first position but the maths works out such that it appears to,since the parabola appears somewhere else on a graph etc etc. What you should gather from this example is that watching out for negative signs and making an appropriate assumption will save you 100000000000000 marks in exams
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Joseph Green
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(Original post by EmRep13)
I thought that but since the acceleration is positive, surely the car doesn't double back? Also your last sentence - "summing them together" - what do you mean?
(Original post by Zoeva123)
When you’re choosing which equation to use you should list out what the question has already given you.
Like:

S=
U=
V=
A=
T=

Your question will always give 3 of the above, the other one is the one you’re calculating and the last one is irrelevant.

So for part a) you’re given s,u and v. You need to calculate a. The irrelevant one is t so you find an equation that does not involve t. The only one is v^2= u^2 +2as which is the accurate formula to be used.

For part b) you have all of the components so you can really use any equation to calculate the time but since there is a possibility of getting part a) incorrect you should choose an equation that doesn’t involve acceleration. ( you said the book used v=u+at but personally I would use s=1/2(u+v)t

This method helps answer any suvat question for a level and gcse maths.
Hope this helps and if you’ve got any other questions let me know x
Zoeva also makes the very valuable point that I forgot about which is: don't forget that if you have to use a previous answer, the value may be incorrect and thus your final answer will be incorrect. You'll still get marks for the method and possible error carried forward marks but it's pretty unlikely that you'll get full marks if you have an incorrect part a)
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EmRep13
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(Original post by Joseph Green)
Summing together is just a fancy way of saying adding. And like I said it's really odd to think about but remember the minus sign makes all the difference here when it comes to drawing the parabola and extrapolating results from it. It's true that the acceleration is positive and it doesn't double back on the first position but the maths works out such that it appears to,since the parabola appears somewhere else on a graph etc etc. What you should gather from this example is that watching out for negative signs and making an appropriate assumption will save you 100000000000000 marks in exams
Yeah I know! I just don't understand at what point you'd add two answers together when doing the quadratic thing? Ah - so the maths gives you two answers but in reality only one is right, unless you're using an situation with deacceleration where it would eventually double back.
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Joseph Green
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(Original post by EmRep13)
Yeah I know! I just don't understand at what point you'd add two answers together when doing the quadratic thing? Ah - so the maths gives you two answers but in reality only one is right, unless you're using an situation with deacceleration where it would eventually double back.
Ohhhhh sorry I believe I wasn't clear. I was assuming for the second part you did 2 calculations. One for s = 100 and one for s = 300 and then added the times together. I forgot you can also do s = 400 😅😅😅😅😅😅
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EmRep13
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(Original post by Zoeva123)
When you’re choosing which equation to use you should list out what the question has already given you.
Like:

S=
U=
V=
A=
T=

Your question will always give 3 of the above, the other one is the one you’re calculating and the last one is irrelevant.

So for part a) you’re given s,u and v. You need to calculate a. The irrelevant one is t so you find an equation that does not involve t. The only one is v^2= u^2 +2as which is the accurate formula to be used.

For part b) you have all of the components so you can really use any equation to calculate the time but since there is a possibility of getting part a) incorrect you should choose an equation that doesn’t involve acceleration. ( you said the book used v=u+at but personally I would use s=1/2(u+v)t

This method helps answer any suvat question for a level and gcse maths.
Hope this helps and if you’ve got any other questions let me know x
Thank you!
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Zoeva123
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(Original post by EmRep13)
Yeah I know! I just don't understand at what point you'd add two answers together when doing the quadratic thing? Ah - so the maths gives you two answers but in reality only one is right, unless you're using an situation with deacceleration where it would eventually double back.
The maths should only actually give you one answer if your method is correct. The car is moving in a straight line so the velocity does not change hence is always positive. If the direction changes you get positive and negative which can then give you 2 answers.
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Joseph Green
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(Original post by Zoeva123)
The maths should only actually give you one answer if your method is correct. The car is moving in a straight line so the velocity does not change hence is always positive. If the direction changes you get positive and negative which can then give you 2 answers.
You aren't wrong about this however the situation with s = vt - 1/2at^2 is the fact that there is a minus sign in there. If you were to form a quadratic using the above equation you'd have 2 positive roots which in this case represent time. This is obviously wrong. Even if acceleration is positive, the minus sign changes the overall placement of the quadratic on a graph which makes things appear wrong. Below is the quadratic formed using that equation
Name:  image-7cb8ae0c-6f2f-441f-b469-d0e43a0f9e048510388185452085251-compressed.jpg.jpeg
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Joseph Green
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(Original post by EmRep13)
Yeah - I was going to type it out for context but felt a bit lazy . Guess I have to now!

A car is travelling along a straight horizontal road with constant acceleration. The car passes over three consecutive points A, B and C where AB = 100m and BC = 300m. The speed of the car at B is 14m/s and the speed of the car at C is 20m/s. Find:

a) The acceleration of the car
b) the time taken for the car to travel from A to C

The answers are 0.34ms^-2 and 25.5 s (3sf) respectively. Typed out because of aforementioned laziness. Thanks!
What was the value of u that they worked out?
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EmRep13
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(Original post by Joseph Green)
What was the value of u that they worked out?
+_ sqrt 128 or +- 8 sqrt 3 and then they decided to use +8sqrt3 assuming the car didn't accelerate reversing backwards
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Joseph Green
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(Original post by EmRep13)
+_ sqrt 128 or +- 8 sqrt 3 and then they decided to use +8sqrt3 assuming the car didn't accelerate reversing backwards
Using the value that they got for u and plugging it into the s = ut + 1/2at^2 you get the same answer and I'll show you. Name:  image-dd2d7ab1-527b-4ec2-b0d6-4c88c3a857b45551799996405536845-compressed.jpg.jpeg
Views: 7
Size:  70.0 KB

The blue quadratic is the s = ut + 1/2at^2 formula and the red is the s = vt - 1/2at^2.

As you can see, the blue one has a positive and negative root, which represent the time solution we are after. The red has 2 positive roots which also both represent the time solutions. The difference is that the red one quite literally creates a false positive. I've also found the intersection between these 2 graphs to show you that the answer is indeed that 25.5 and using either method that involves the quadratic suvat equations will result in the right solution provided you choose the right answer.

I hope this helps a little
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EmRep13
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(Original post by Joseph Green)
Using the value that they got for u and plugging it into the s = ut + 1/2at^2 you get the same answer and I'll show you. Name:  image-dd2d7ab1-527b-4ec2-b0d6-4c88c3a857b45551799996405536845-compressed.jpg.jpeg
Views: 7
Size:  70.0 KB

The blue quadratic is the s = ut + 1/2at^2 formula and the red is the s = vt - 1/2at^2.

As you can see, the blue one has a positive and negative root, which represent the time solution we are after. The red has 2 positive roots which also both represent the time solutions. The difference is that the red one quite literally creates a false positive. I've also found the intersection between these 2 graphs to show you that the answer is indeed that 25.5 and using either method that involves the quadratic suvat equations will result in the right solution provided you choose the right answer.

I hope this helps a little
Thanks so much for your help! It really cleared so much up! Thanks for your help too Zoeva123
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RogerOxon
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(Original post by EmRep13)
I have a question where I have s,v and a. I need to find t. I used s = vt - 1/2at^2 and got two answers. The book used v^2 = u" +2as to find u then used v= u + at to find t, so got one answer, which I also got as one of my two. My question is - are there certain times you use certain equations? Like obviously, if it said what are the possible value(s) you'd use the one that gives a quadratic? If not - then why did I get two answers? The situation in question is about a car travelling from A to C in a certain time, accelerating.

Edit: Grammar
The book should have got two answers for u (plus and minus the square root), and discounted one.
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EmRep13
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(Original post by RogerOxon)
The book should have got two answers for u (plus and minus the square root), and discounted one.
It does! It explains that we use plus the square root!
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