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# SUVAT equations (velocity, time etc) watch

1. Hey,
I am looking at this question, it is on the WJEC board, on SUVAT equations. It says the following:

A car accelerates for four seconds at a constant rate. Over the four seconds it travels 104m. It covered 58m in the last two seconds. Find the speed at which the car was travelling before it started to accelerate and the rate of acceleration (in ms-2).

I am basically quite confused at how I am meant to find the answer using one of the four equations when I only have two values?

Any help would be REALLY appreciated!

Thank youuuu!!
2. (Original post by alicebrr)
Hey,
I am looking at this question, it is on the WJEC board, on SUVAT equations. It says the following:

A car accelerates for four seconds at a constant rate. Over the four seconds it travels 104m. It covered 58m in the last two seconds. Find the speed at which the car was travelling before it started to accelerate and the rate of acceleration (in ms-2).

I am basically quite confused at how I am meant to find the answer using one of the four equations when I only have two values?

Any help would be REALLY appreciated!

Thank youuuu!!
You can set up two equations to solve for two unknowns.
You have time and distance, and you need to find acceleration and initial speed. What equation do you need?
This is a nice question actually, a little tricky
3. (Original post by EricPiphany)
You can set up two equations to solve for two unknowns.
You have time and distance, and you need to find acceleration and initial speed. What equation do you need?
That's the issue, I just don't know where to begin!
It's one of these:

v=u+at

s=1/2(u+v)t

s=ut+1/2at^2

v^2=u^2+2as
4. (Original post by alicebrr)
That's the issue, I just don't know where to begin!
It's one of these:

v=u+at

s=1/2(u+v)t

s=ut+1/2at^2

v^2=u^2+2as
Choose one that contains the variables you know, and the ones you are looking for.
5. (Original post by EricPiphany)
Choose one that contains the variables you know, and the ones you are looking for.
So that's s=ut+1/2at^2, right?
And do I then have to use another equation to solve simultaneously or something? Because I got 208=4u+16a^2, which doesn't give me either value, obviously! I was fine up until this question, and the wording just threw me! Haha!
6. (Original post by alicebrr)
So that's s=ut+1/2at^2, right?
And do I then have to use another equation to solve simultaneously or something? Because I got 208=4u+16a^2, which doesn't give me either value, obviously! I was fine up until this question, and the wording just threw me! Haha!
Remember, you're given "two journeys", one is the entire journey and the other is the last half of the journey. So you can use the same equation with the same variables but different coefficients, so to speak, to find two equations containing u and a, allowing you to solve for those two simultaneously.
7. (Original post by alicebrr)
So that's s=ut+1/2at^2, right?
And do I then have to use another equation to solve simultaneously or something? Because I got 208=4u+16a^2, which doesn't give me either value, obviously! I was fine up until this question, and the wording just threw me! Haha!
Shouldn't that be 104 = 4u - 8a or 26 = u - 2a?
8. (Original post by EricPiphany)
Shouldn't that be 104 = 4u - 8a or 26 = u - 2a?
Ah, yeah...but how is it negative? I thought it would have been 104=4u+8a?
9. (Original post by Zacken)
Remember, you're given "two journeys", one is the entire journey and the other is the last half of the journey. So you can use the same equation with the same variables but different coefficients, so to speak, to find two equations containing u and a, allowing you to solve for those two simultaneously.
Slight trick here, you have to update u using a second suvat equation.
10. (Original post by alicebrr)
Ah, yeah...but how is it negative? I thought it would have been 104=4u+8a?
'a' is positive. It has a negative coefficient in the equation. You haven't found 'a' just yet
11. (Original post by EricPiphany)
'a' is positive. It has a negative coefficient in the equation. You haven't found 'a' just yet
I am so confused! Ha! I think I will go and ask my teacher, as I have the first equation, but you lost me after that! Thankyou for all of your help though!
12. (Original post by alicebrr)
I am so confused! Ha! I think I will go and ask my teacher, as I have the first equation, but you lost me after that! Thankyou for all of your help though!
No problem.
13. I'm still at GCSE level, and I liked this question - allowed me to think :P but it seemed straightforward when you explore the SUVAT equations.
Somehow, I got two simultaneous equations and I'm on the right track :O.

Whenever you get the actual answer, please let me know so I can see if my answer is also right or if it's wrong .
14. (Original post by Chittesh14)
I'm still at GCSE level, and I liked this question - allowed me to think :P but it seemed straightforward when you explore the SUVAT equations.
Somehow, I got two simultaneous equations and I'm on the right track :O.

Whenever you get the actual answer, please let me know so I can see if my answer is also right or if it's wrong .
15. (Original post by alicebrr)
Hey,
I am looking at this question, it is on the WJEC board, on SUVAT equations. It says the following:

A car accelerates for four seconds at a constant rate. Over the four seconds it travels 104m. It covered 58m in the last two seconds. Find the speed at which the car was travelling before it started to accelerate and the rate of acceleration (in ms-2).

I am basically quite confused at how I am meant to find the answer using one of the four equations when I only have two values?

Any help would be REALLY appreciated!

Thank youuuu!!
A far better way is to use calculus, which has the key advantage that it works even when acceleration is not constant. A good introduction is Keisler:
https://www.math.wisc.edu/~keisler/k...c-12-20-15.pdf. Read at least Chapters 1-4, then you can solve your problem simply by using the fact that acceleration is the rate of change of velocity (a=dv/dt) and velocity is the rate of change of distance (v=ds/dt).
16. (Original post by constellarknight)
A far better way is to use calculus, which has the key advantage that it works even when acceleration is not constant. A good introduction is Keisler:
https://www.math.wisc.edu/~keisler/k...c-12-20-15.pdf. Read at least Chapters 1-4, then you can solve your problem simply by using the fact that acceleration is the rate of change of velocity (a=dv/dt) and velocity is the rate of change of distance (v=ds/dt).
LOLOLOLOL good one

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