SUVAT equations?!
I'm trying to calculate the acceleration of something travelling with a certain speed, and then stopping with a speed of 0m/s. I am assuming the object decelerates at a constant rate. Which suvat equation do I use?
I used a = (v - u)/t and got a reasonable answer.
But I also noticed that a = (2s-2ut)/(t^2) and that a = (v^2 - u^2)/2s
All give me different answers - the second one I mentionned gives me double the deceleration given by the first equation, and the 3rd gives something nearly 100 times greater!! Which do I use?
I used a = (v - u)/t and got a reasonable answer.
But I also noticed that a = (2s-2ut)/(t^2) and that a = (v^2 - u^2)/2s
All give me different answers - the second one I mentionned gives me double the deceleration given by the first equation, and the 3rd gives something nearly 100 times greater!! Which do I use?
Scroll to see replies
Original post by the bear
They should all give the correct answer if the acceleration is constant
A car has mass = 820kg, initial velocity is (u=) 50km/h (=13.8889m/s); then brakes are applied for a distance of (s=) 1.8cm (=0.018m) and then the car stops (final velocity, v = 0m/s). I have to work out the force required for this car to stop in a distance of 1.8cm (very small!!!). I thought I should use F=ma, and work out the acceleration (deceleration) using suvat...but nooo it doesn't work
Some guidance (not the actual answers/every single step) would be much appreciated, as it's for an assignment
Original post by PhysicsGal
A car has mass = 820kg, initial velocity is (u=) 50km/h (=13.8889m/s); then brakes are applied for a distance of (s=) 1.8cm (=0.018m) and then the car stops (final velocity, v = 0m/s). I have to work out the force required for this car to stop in a distance of 1.8cm (very small!!!). I thought I should use F=ma, and work out the acceleration (deceleration) using suvat...but nooo it doesn't work
Some guidance (not the actual answers/every single step) would be much appreciated, as it's for an assignment
Some guidance (not the actual answers/every single step) would be much appreciated, as it's for an assignment
Original post by Voyageuse
What are the values you know? So u=initial velocity, v=0. What other values are you given?
^^^^^^^
Also, I worked out that the car travels the short distance of 0.018m in 0.25s (a quarter of a second)...and the car has a weight of 8044.2N (1 d.p.) but not sure if I really need the weight :/
Original post by PhysicsGal
A car has mass = 820kg, initial velocity is (u=) 50km/h (=13.8889m/s); then brakes are applied for a distance of (s=) 1.8cm (=0.018m) and then the car stops (final velocity, v = 0m/s). I have to work out the force required for this car to stop in a distance of 1.8cm (very small!!!). I thought I should use F=ma, and work out the acceleration (deceleration) using suvat...but nooo it doesn't work
Some guidance (not the actual answers/every single step) would be much appreciated, as it's for an assignment
Some guidance (not the actual answers/every single step) would be much appreciated, as it's for an assignment
This should get you the right answer... could you post your working here perhaps?
Original post by justinawe
This should get you the right answer... could you post your working here perhaps?
Sure (kinda posted a little in the above few posts) - it's just when I'm trying to calculate acceleration using suvat it doesn't work. I thought I'd use 3 suvat equations just to check my acceleration is the same with each...but they give different answers :/
s = 0.018m (in which car goes from velocity u to velocity v)
u = 50km/h = 13.8888888888889m/s (using the calc. value not the rounded one)
v = 0m/s
a = UNKNOWN
t = 0.25s (in which car goes from velocity u to velocity v)
Then use F=ma =820a to work out F (force required for car to go from velocity u to velocity v)
a = (v-u)/t = 13.88888888889/0.25 = 55.555555555556 m/s^2
a = [2(s-ut)]/(t^2) = (0.036-3.472222222)/0.0625 = 110.53511111 m/s^2
a = (v^2 - u^2)/2s = [0 - (13.888889^2)]/(0.036) = 5358.37... m/s^2
Pfft :/
Original post by shwareb
I would use F = ma but also the equation v^2= u^2 + 2as
1. v^2= u^2 + 2as therefore a= (v^2-u^2)/2s
2. a=( 0-13.9)/(2*0.018)ms^-2 = -386.1 ms^-2
3. F=ma, F=820*386.1= 316602 = 320 000 N (Answer in with two significant figures)
1. v^2= u^2 + 2as therefore a= (v^2-u^2)/2s
2. a=( 0-13.9)/(2*0.018)ms^-2 = -386.1 ms^-2
3. F=ma, F=820*386.1= 316602 = 320 000 N (Answer in with two significant figures)
How come you would not use the other suvat equations?
Thanks
I get a very large deceleration (-5358.4m/s^2) using two different equations:
1) u=13.8889, v=0, s=0.018, a=?
v^2=u^2+2as 0=13.8889^2+2xax0.018 -13.8889^2/0.036=a=5358.4m/s^2
Are you sure your value for t is right? I used s=((v+u)t)/2 and got 0.00259s
2) u=13.8889, v=0, t=0.00259, a=?
v=u+at 0=13.8889+0.00259a -13.8889/(0.036/13.8889)=a=5358.4m/s^2
1) u=13.8889, v=0, s=0.018, a=?
v^2=u^2+2as 0=13.8889^2+2xax0.018 -13.8889^2/0.036=a=5358.4m/s^2
Are you sure your value for t is right? I used s=((v+u)t)/2 and got 0.00259s
2) u=13.8889, v=0, t=0.00259, a=?
v=u+at 0=13.8889+0.00259a -13.8889/(0.036/13.8889)=a=5358.4m/s^2
Original post by PhysicsGal
How come you would not use the other suvat equations?
Thanks
Thanks
Well I just really picked this one when I saw the question!
The four SUVAT equations are:
1. s= ut + (1/2)at^2 But I´m not given the time it takes for the car to stop and therefore I will not use this equation.
2. v= u + at Same thing here; I´m not given the time it takes for the car to stop and therefore I will not use this equation.
3. s = ((v+u)/2) * t Same thing here; I´m not given the time it takes for the car to stop and therefore I will not use this equation. This equation also doesn´t give us the acceleration so really this is out of the question.
4. v^2 = u^2 + 2as... THIS IS THE EQUATION WOOHOO!!!
In the question we are given the initial velocity (u= 50 km/h), we are given the final velocity (v = 0 km/h) and we are given the distance travelled ( s= 0.018m). Therefore I am able to calculate the acceleration.
(edited 11 years ago)
Original post by Voyageuse
I get a very large deceleration (-5358.4m/s^2) using two different equations:
1) u=13.8889, v=0, s=0.018, a=?
v^2=u^2+2as 0=13.8889^2+2xax0.018 -13.8889^2/0.036=a=5358.4m/s^2
Are you sure your value for t is right? I used s=((v+u)t)/2 and got 0.00259s
2) u=13.8889, v=0, t=0.00259, a=?
v=u+at 0=13.8889+0.00259a -13.8889/(0.036/13.8889)=a=5358.4m/s^2
1) u=13.8889, v=0, s=0.018, a=?
v^2=u^2+2as 0=13.8889^2+2xax0.018 -13.8889^2/0.036=a=5358.4m/s^2
Are you sure your value for t is right? I used s=((v+u)t)/2 and got 0.00259s
2) u=13.8889, v=0, t=0.00259, a=?
v=u+at 0=13.8889+0.00259a -13.8889/(0.036/13.8889)=a=5358.4m/s^2
I used time = distance/speed ....now I bet that's wrong, but aren't sure why :/ I remember you can't use that when using suvat but why?
I figured that if the car travels 13.8889 metres every second, then to travel 0.018 metres it takes 13.8889*0.018 = 0.25 seconds
Anyway, so using your value of t should give the same 'a' throughout the equations right?
Original post by PhysicsGal
I used time = distance/speed ....now I bet that's wrong, but aren't sure why :/ I remember you can't use that when using suvat but why?
I figured that if the car travels 13.8889 metres every second, then to travel 0.018 metres it takes 13.8889*0.018 = 0.25 seconds
Anyway, so using your value of t should give the same 'a' throughout the equations right?
I figured that if the car travels 13.8889 metres every second, then to travel 0.018 metres it takes 13.8889*0.018 = 0.25 seconds
Anyway, so using your value of t should give the same 'a' throughout the equations right?
because velocity isn't constant
use since that a is the only unknown in that equation
Original post by PhysicsGal
Sure (kinda posted a little in the above few posts) - it's just when I'm trying to calculate acceleration using suvat it doesn't work. I thought I'd use 3 suvat equations just to check my acceleration is the same with each...but they give different answers :/
s = 0.018m (in which car goes from velocity u to velocity v)
u = 50km/h = 13.8888888888889m/s (using the calc. value not the rounded one)
v = 0m/s
a = UNKNOWN
t = 0.25s (in which car goes from velocity u to velocity v)
Then use F=ma =820a to work out F (force required for car to go from velocity u to velocity v)
a = (v-u)/t = 13.88888888889/0.25 = 55.555555555556 m/s^2
a = [2(s-ut)]/(t^2) = (0.036-3.472222222)/0.0625 = 110.53511111 m/s^2
s = 0.018m (in which car goes from velocity u to velocity v)
u = 50km/h = 13.8888888888889m/s (using the calc. value not the rounded one)
v = 0m/s
a = UNKNOWN
t = 0.25s (in which car goes from velocity u to velocity v)
Then use F=ma =820a to work out F (force required for car to go from velocity u to velocity v)
a = (v-u)/t = 13.88888888889/0.25 = 55.555555555556 m/s^2
a = [2(s-ut)]/(t^2) = (0.036-3.472222222)/0.0625 = 110.53511111 m/s^2
The bold number is wrong
Right sorry I forgot to take the initial velocity squared 
"I would use F = ma but also the equation v^2= u^2 + 2as
1. v^2= u^2 + 2as therefore a= (v^2-u^2)/2s
2. a=( 0-13.9^2)/(2*0.018)ms^-2 = -5367 ms^-2
3. F=ma, F=820*5367= 4400940 = 4 400 000 N (Answer in with two significant figures) "
Correct answer is 4 400 000 N
Don´t bother calculating the time when you can use this equation: v^2= u^2 + 2as
"I would use F = ma but also the equation v^2= u^2 + 2as
1. v^2= u^2 + 2as therefore a= (v^2-u^2)/2s
2. a=( 0-13.9^2)/(2*0.018)ms^-2 = -5367 ms^-2
3. F=ma, F=820*5367= 4400940 = 4 400 000 N (Answer in with two significant figures) "
Correct answer is 4 400 000 N
Don´t bother calculating the time when you can use this equation: v^2= u^2 + 2as
(edited 11 years ago)
You can't use speed=distance/time if the object is not moving at a constant speed (this one is decelerating). Using t=0.00259 gives the same value of a using different equations.
I think you forgot to square your u in part 1 here? a=(v^2-u^2)/2s=(0-13.8889^2)/0.036=-5358.37...
Original post by shwareb
I would use F = ma but also the equation v^2= u^2 + 2as
1. v^2= u^2 + 2as therefore a= (v^2-u^2)/2s
2. a=( 0-13.9)/(2*0.018)ms^-2 = -386.1 ms^-2
3. F=ma, F=820*386.1= 316602 = 320 000 N (Answer in with two significant figures)
1. v^2= u^2 + 2as therefore a= (v^2-u^2)/2s
2. a=( 0-13.9)/(2*0.018)ms^-2 = -386.1 ms^-2
3. F=ma, F=820*386.1= 316602 = 320 000 N (Answer in with two significant figures)
I think you forgot to square your u in part 1 here? a=(v^2-u^2)/2s=(0-13.8889^2)/0.036=-5358.37...
Original post by shwareb
Well I just really picked this one when I saw the question!
The four SUVAT equations are:
1. s= ut + (1/2)at^2 But I´m not given the time it takes for the car to stop and therefore I will not use this equation.
2. v= u + at Same thing here; I´m not given the time it takes for the car to stop and therefore I will not use this equation.
3. s = ((v+u)/2) * t Same thing here; I´m not given the time it takes for the car to stop and therefore I will not use this equation. This equation also doesn´t give us the acceleration so really this is out of the question.
4. v^2 = u^2 + 2as... THIS IS THE EQUATION WOOHOO!!!
In the question we are given the initial velocity (u= 50 km/h), we are given the final velocity (v = 0 km/h) and we are given the distance travelled ( s= 0.018m). Therefore I am able to calculate the velocity.
The four SUVAT equations are:
1. s= ut + (1/2)at^2 But I´m not given the time it takes for the car to stop and therefore I will not use this equation.
2. v= u + at Same thing here; I´m not given the time it takes for the car to stop and therefore I will not use this equation.
3. s = ((v+u)/2) * t Same thing here; I´m not given the time it takes for the car to stop and therefore I will not use this equation. This equation also doesn´t give us the acceleration so really this is out of the question.
4. v^2 = u^2 + 2as... THIS IS THE EQUATION WOOHOO!!!
In the question we are given the initial velocity (u= 50 km/h), we are given the final velocity (v = 0 km/h) and we are given the distance travelled ( s= 0.018m). Therefore I am able to calculate the velocity.
Original post by justinawe
because velocity isn't constant
Ahh thank you guys!!
So my intial calculation of time screwed it up, but with the new time using suvat, I have a consistent acceleration regardless of which suvat equation I use...makes sense! Thanks
Edit: Ran out of rep
but 
(edited 11 years ago)
Original post by Voyageuse
You can't use speed=distance/time if the object is not moving at a constant speed (this one is decelerating). Using t=0.00259 gives the same value of a using different equations.
I think you forgot to square your u in part 1 here? a=(v^2-u^2)/2s=(0-13.8889^2)/0.036=-5358.37...
I think you forgot to square your u in part 1 here? a=(v^2-u^2)/2s=(0-13.8889^2)/0.036=-5358.37...
Yeah, I corrected myself
. Look one post above you.Original post by shwareb
Right sorry I forgot to take the initial velocity squared
"I would use F = ma but also the equation v^2= u^2 + 2as
1. v^2= u^2 + 2as therefore a= (v^2-u^2)/2s
2. a=( 0-13.9^2)/(2*0.018)ms^-2 = -5367 ms^-2
3. F=ma, F=820*5367= 4400940 = 4 400 000 N (Answer in with two significant figures) "
Correct answer is 4 400 000 N
Don´t bother calculating the time when you can use this equation: v^2= u^2 + 2as
"I would use F = ma but also the equation v^2= u^2 + 2as
1. v^2= u^2 + 2as therefore a= (v^2-u^2)/2s
2. a=( 0-13.9^2)/(2*0.018)ms^-2 = -5367 ms^-2
3. F=ma, F=820*5367= 4400940 = 4 400 000 N (Answer in with two significant figures) "
Correct answer is 4 400 000 N
Don´t bother calculating the time when you can use this equation: v^2= u^2 + 2as
I agree with this answer ^^ F=820*5358.37=4393863.4 N
Original post by Voyageuse
I agree with this answer ^^ F=820*5358.37=4393863.4 N
I think you used u=13.8889 m/s for initial velocity were as I rounded it up to u= 13.9 m/s. That is why we get different answers!
You are more accurate however
Quick Reply
Related discussions
- Mechanics A Level Maths Problem
- Physics GCSE
- A Level Physics Mechanics HELPP!!!
- m1 mechanics as vectors 2018 internation edexcel q6
- A-level Maths Mechanics: SUVAT Question HELP
- A Level Maths mechanics help (vectors)
- I don't like suvat equations, can I sketch graphs instead?
- Maths based project
- A level maths mechanics question (1)
- Help maths super urgent 😭
- help me with projectiles before i throw my laptop out of a window
- pulley
- pulleys-mechanics, pls help i cant solve the whole question
- Fsmq ocr additional maths 23 jun 2023
- Help mechanics urgent
- A Level Mechanics.
- projectile maths question alevel
- Can anyone help with this SUVAT A-level question?
- Math question
- Isaac Physics: Will they collide
Latest
Last reply 2 minutes ago
Official London School of Economics and Political Science 2024 Applicant ThreadLast reply 5 minutes ago
Government Social Research - Research Officer Scheme 2024Last reply 9 minutes ago
Official Cambridge Postgraduate Applicants 2024 ThreadLast reply 9 minutes ago
choosing a medical school (Leicester, St Andrews, Nottingham)Last reply 13 minutes ago
Why do some girls take pleasure in going after other girls boyfriends?Posted 14 minutes ago
LSE undergraduate accommodationLast reply 16 minutes ago
Can someone please mark my AIC essay (my teacher is such an unreliable marker)Last reply 16 minutes ago
JK Rowling in ‘arrest me’ challenge over hate crime lawLast reply 17 minutes ago
CPS Pupillage/Legal Trainee Scheme 2024 (2025 Start)Last reply 18 minutes ago
Best Accommodation for Med Students at University of Plymouth?Last reply 18 minutes ago
Official University of Warwick Offer Holders Thread for 2024 entryLast reply 21 minutes ago
Bae systems accommodation barrowLast reply 24 minutes ago
Official University of Edinburgh Applicant Thread for 2024Last reply 25 minutes ago
University of Liverpool veterinary work experience form.
