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# SUVAT equations?! Watch

1. 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?
2. What's the actual question?
3. They should all give the correct answer if the acceleration is constant
4. What are the values you know? So u=initial velocity, v=0. What other values are you given?
5. (Original post by lmsavk)
What's the actual question?
(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
6. (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
(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 :/
7. (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
This should get you the right answer... could you post your working here perhaps?
8. 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)
9. (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 :/
10. (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)

How come you would not use the other suvat equations?

Thanks
11. 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
12. (Original post by PhysicsGal)
How come you would not use the other suvat equations?

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.

13. (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

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?
14. (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?
because velocity isn't constant

use since that a is the only unknown in that equation
15. (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
The bold number is wrong
16. 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
17. 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.
(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)
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...
18. (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.

(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
19. (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...
Yeah, I corrected myself . Look one post above you.
20. (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 agree with this answer ^^ F=820*5358.37=4393863.4 N

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