# Which equation?

A car drives along a dual carriageway at 96 km h−1. The driver sees a speed limit sign approaching and applies the brakes to decelerate to a speed of 48 km h−1. Given that the deceleration is 2.68 m s−2 find the distance over which the brakes are applied.

Formula I think to use is to basratardise this formula which uses potential energy to find the distance

Which is

Ek = ½ mv2

DeltaEg = mg Delta h

Which equates to

mg Delta h = ½ mv2

which would change to

delta h = v²/2g

I would change h for d=distance
V² would equal 48 km h−1 converted to m sˉ¹
g would I would change to a for acceleration (deceleration)= 2.68 m s−2

so it would be

delta d = v²/2a

can I do this? My thinking is that energies are conserved so it would be ok to do this?

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Doc Al
Mentor
That's pretty close. But rather than try to force fit some equation meant for a different situation, derive a new one. If you want to use energy methods, how about considering the work done by the braking force? Otherwise, just use kinematics--after all, you're given the acceleration.

i'm doing ou study and must use equations given in the course books nearly all the equations accept the one i've shown require mass to be known?

Doc Al
Mentor
You don't need to know the mass for either the energy method or the kinematics solution. (Just call the mass 'm' and proceed.)

If you listen to Doc Al(and I would if I were you) I'd stay with kinematics equations.

The equation you have works fine (apart from needing to remember that $$v^{2}$$ is in fact $$v_{f}^{2} - v_{0}^{2}$$ and you should always state the direction of positive motion in your answer.

i thank you for your replies the course is foundation level and there is now mention of kinematics, and it has not showed me how to just use m in formulas and that last thread just confused me even more. I'm OK in use of equations and my maths is not to bad but i must only work with the the equations in the books and not other outside sources (i think that's for later courses should I pursue more). So you see the equations i stated are the only one that does not include mass.

As I said, your equation is correct, just the notation is a bit iffy(and the derivation is not completely sound, but it'll do if you're doing a foundation course.

As long as you know what velocity you are using as $$v$$ then you should have no problems.

The reason believe i can use the equation is this

A diver who dives off the famous cliffs at Acapulco in Mexico hits the
water at about 25 m s −1 . Calculate the height of the cliff.

the diver has a potential energy of 9.8 m/s/s which equates to 2.68 m/s/s in the question the final speed is 1.3 m/s which equates to 25 m/s. Height is is linear measurement so is distance. potential energy can become kinetic energy which is the car moving at its initial speed. does this make sense?

The reason believe i can use the equation is this

A diver who dives off the famous cliffs at Acapulco in Mexico hits the
water at about 25 m s −1 . Calculate the height of the cliff.

the diver has a potential energy of 9.8 m/s/s which equates to 2.68 m/s/s in the question the final speed is 1.3 m/s which equates to 25 m/s. Height is is linear measurement so is distance. potential energy can become kinetic energy which is the car moving at its initial speed. does this make sense?
Not to me, this doesn't:
1)You've stated potential energy as an acceleration
2) how can $$1.3ms^{-1} = 25ms^{-1}$$ ?
3)If I were you I wouldn't say energy 'becomes' other energy, say 'converted to...' or 'transferred in the system as...'
4)Why are we talking about Mexico?

the acceleration before he braked was constant so what i am saying is that because it is constant it is as if the energy is like then potential energy of the the diver on the top of the cliff there is no change until he breaks? So could i zero that speed (not use it in the the equation) In regard to the 1.3 m/s = 25m/s the 1.3 is the final speed of the car when it has slowed down which was the final speed of the diver as he hits the water. And where talking about Mexico because this was a question asked in the course book to see if i understood the equation i was using?:-)

My Answer for the breaking vehicle using this is 31.5m if you did it doing they way you would do it, is the answer the same if it is just concur yes or no i do not need to no your workings as i need to find this out for myself, if however i am correct then i am willing to drop the marks that might have been given, but i will have it explained where i went wrong.

Mentally just now I get the answer to be ~100m, but it will be slightly less than this.

Don't take this answer as final, it could very well be wrong, as it is only me estimating a square root.

now somewhere along the way I've had an answer around there and i cannot remember how so i'll have to scroll back through everything I've done cos the answer i got is well short of that?

jack action
Gold Member
With your equation, you will get the answer like everyone else is telling you. But I think it is more important to understand where this equation comes from.

You have the initial kinetic energy of the vehicle:

$$E_{ko}=\frac{1}{2} mv_o^{2}$$

You have the final kinetic energy of the vehicle:

$$E_{kf}=\frac{1}{2} mv_f^{2}$$

The work done by the brakes is defined as a force multiplies by the distance on which it is applied. In this case the force is equal to the mass times the acceleration:

$$W_b=F_b\Delta s=ma\Delta s$$

So based on the concept of conservation of energy, if you remove the work done by the brakes from the initial kinetic energy, you should get the final kinetic energy of the vehicle:

$$E_{kf}=E_{ko}-W_b$$

Do the appropriate transformations and you will get the equation you need to find the distance $$\Delta s$$.

The same equations will apply to your diver problem.

I personally thought that the mechanical approach was simpler on this occasion...even though some people insist on being 'sophisticated' about certain problems; you need to know which approach is better for each instance.

I know you stated that you have to use those equations, but imo that is a bad move by the examining body; where in life is an employer going to say "You got the perfect result, but the direction of approach was wrong" - never, as long as it gets the job done.

Now, kinematics. =]

If you simply use the equation $$v_{f}^{2} = v_{0}^{2} + 2as$$ you can rearrange to find s easily; giving the form:

$$s = \frac{v_{f}^{2} - v_{0}^{2}}{2a}$$ and sub in your values.

After sitting here for nearly 6 hours now i have finally cottoned on to they way to do it, i probably will not be spot on with my thinking and equation not totally scientifically correct but using my initial equation and looking at the adjustments that you have all suggested and going on a website that explained it through kinematics then cross reference's i have managed to get the correct answer (about 100m). I have to stick to the teaching method in the books and it's foundation level. So i am prepared to go with that and see how the lecturer marks it and learn from that. so i thank you all for your help and i will probably be needing your help in the future:_)