Can anyone help me with this maths please?

[sealsandbeachballs]

DD has worked out what speed a car was going based on it's skid marks on a rainy day. It was over the speed limit.

How do we then do the maths to work out if the cars would still have crashed if going at the speed limit?

Depends on what you know?

Point of origin, distance etc?

It says cars traveling on an urban road with a 30 mph speed limit on a rainy day. The driver fled the scene, the pedestrian ended up unconscious in hospital.

then it says something about -6.80 ms-2 Accelaration (lower)
-7.01 ms-2 upper acceleration

(maths is not my strong point and I have no clue!)

oh and the calculation

0= u squared + 2x (-6.80ms-2squared)

u = initial speed

Start by drawing a new diagram of the two cars and label all the known things - so for car A you have a new starting speed, final speed is still zero.

Drawing a diagram often really helps students if they are struggling to find a starting point. To help more than that we need more information.
Is she using the suvat equations?
What do you know about car B?

Sorry, X-post!

So she can use the same equation, but this time she knows the starting velocity and needs to calculate the distance of the skid.

sorry have just checked and they are not interested in the other car. The car traveling 40 mph hit a pedestrian. They want to know if he was doing 30 not 40 if he would have still hit the pedestrian.

I have no information on the pedestrian other than he was near a pub on an urban road.

How do we work out if he would have hit the pedestrian with no information on the pedestrian [thick emotion]

Do you know how far apart the pedestrian was from the start of the skid mark?

I hate typing equations, so I've written it instead.

no it doesn't say!!

Can you explain to me how the equation works?

if the acceleration was -6.80 lower and -7.01 upper value is that the change in velocity?

we know that the car skidded 17m 40cm
how does she then work out the speed (she has worked it out but can't remember how)

and how do we then work out how far the car would have traveled at 30mph (she has worked out the answer is just over 13 metres per second but I cannot figure out how)

sorry x post! thank you

So the letters:

s is displacement (you can think of it as distance)
u is initial velocity
v is final velocity
a is acceleration

If you have any three of those things you use the equation to calculate the thing you don't know.

In the initial part of the problem, you had the final velocity, displacement (the length of the skid mark) and acceleration. So you then calculate the missing thing - the initial velocity.

For the second part it is helpful to think of it as a totally new problem. I tend to insist students draw a second diagram to help make the distinction. This time you know the initial velocity, final velocity and acceleration, so you use the equation to calculate the displacement.

Also, can you take a picture of the whole question so I can see how it is worded? It might make a difference to which value of acceleration you need to use.

-6.80ms-2 and -7.01ms-2 are both values for acceleration.

ohhh I get it I think

So dd has put 0 = u2 + 2 x (-6.80 acceleration) x 245 ( she has done joint displacement of both cars but if joint displacement of both cars is 24.5 should she have done x 2.45 not 245 ?)

sorry 24.5 not 245 is what I meant!

I don't understand this upper and lower thing. I'd expect them to assume constant deceleration.

For the first part of the problem the length of the skid was 17.4m so that is the displacement she should use because that is the length travelled while the car was braking.

Though I don't understand what you mean by two cars.

As I understood it there was
a) the real situation with the speeding car, crash and skid marks
b) a theoretical situation with a car travelling at the speed limit, a skid of unknown distance and a potential crash

They have to be treated as two separate problems.

I have got it thanks, managed to work it out from what dd had wrote and what you had wrote.

There were three problems.
One car skids in wet at 40mph and hits man
One car skids in wet at 30mph and work out if he hits the man
another page that i have just found where car 1 hits man, car 2 hits car 1

Thank you I get it now

stealth, to use those equations it does have to be constant acceleration. That's why I'd need to see the wording of the question to determine which value to use.

For instance, the question could be "what is the minimum velocity the car could have been travelling at?", in which case you have to use the lower value.

If the question was "what velocity was the car travelling at? Calculate the percentage error in your answer." You would have to use the mean value for acceleration to find the starting velocity and then use the other values to find the percentage error.

And there are million similar questions which would need you to do something slightly different.

You are welcome seals. Your DD really should ask her teachers if she gets stuck. I personally would rather a student came to me for help (after making a reasonable attempt) than struggling for hours on end at home. The vast majority of teachers I work with feel the same. Tho obviously that isn't always easy for some students.

But she does give two acceleration values
but then the formula for constant acceleration
Makes no sense

The upper limit and lower limit are usually because it's a real problem rather than a theory. So in an actual investigation they couldn't know for certain what the value for acceleration was. So they would have an upper limit and a lower limit of what would be expected for the car in question.

Then in calculating the minimum velocity you give the driver the benefit of the doubt and use the lower limit value.

Also, I swear I am better at explaining this in real life! I've never tried to teach via a forum before - it is so much harder than in person.

It says on the second sheet I later found that if the tests are within 10% of each other to use the low value for the calculations.

She had done near enough all of it but couldn't remember how she had done it and needed help for the final bit.

You had to work out who was responsible for the accident car 1, car 2 or the pedestrian.

right dd has got

Car 1 was

v2 = u2 + 2as
u = 02 + 2 x (-6.80) x 17.4 (distance)
-236.64
15.38 metres per second which I think is roughly 34.4mph?

Is she completely wrong with the equation or on the right track?

Looks about right but my calculator is in school so I can't check the numbers. The only thing is that she has rearranged the equation slightly incorrectly. That is why she ended up with a negative value for u squared, which wouldn't give her any answer of u and (I presume) she had to just ignore the negative sign to be able to get her calculator to give an answer. See photo for a correct version.

There are other correct ways of rearranging the equation and she needs to speak to her teacher about what works best for her. It looks (to me) like she was trying to do two steps in one because it isn't a terribly difficult one and it is usually works fine, but that leads to mistakes when tired or under exam stress.