Equations of Motion (help)

Also, I am pullin my hair out at this one:

The two ends of a train moving with constant acceleration pass a certain point with velocities "u" and "v". Find in terms of "u" and "v" what proportion of the length of the train will have passed the point after a time equal to half that taken by the train to pass it.

Please help, i just can't get this!

Re: Equations of Motion (help)


There are 2 variables in 1 equation. You can't solve for both!

You need to use the information about the starting distance between them.

I will all the fast one car A and the slow one car B.

You know that B.x(0) - A.x(0) = 500 because this is the difference in starting positions, and you know that B.x(f) = A.x(f) because they will be at the same position at the end of the acceleration.

You also know that A.v(f) = B.v(f) because they will be going the same speed at the end.

Using: x = x0 + v0t + 1/2at^2, we write equations for both and set them equal because final positions are equal.

A.x(0) + A.v(0)*t + 1/2at^2 = B.x(0) + B.v(0)*t

Now we can substitute in the difference...

A.v(0)*t + 1/2at^2 = 500 + B.v(0)*t

Now subst in the numbers

(1) 88*t + 1/2at^2 = 500 + 29.33*t

To get the other equation, use v = v0 + at on car A.

A.v(f) = A.v(0) + a*t

Substituing in, we get

(2) 29.33 = 88 + a*t

Now we have 2 eq's and 2 unknowns. Solve (2) for 'a':

a = (29.33 - 88)/t

Substitute that into equation (1), this gives value of t, plug back into eq (2) and that gives you a.

But we know that for all real and positive values of "t" that
0.5at^2 + 58 2/3 t -600 <0
must be satisfied, therefore, from that, can we derive the values of "a" that will satisfy the inequality? The graph must open downwards (if you know what I mean), it must be completely below the x- axis, therefore a<o and the roots are non-real right?



I can't see why they must be going the same speed at the end...they want to AVOID a collision.

Your lettering is odd, could you please rephrase that as s=displacement
u=initial velocity v=final velocity a=acceleration t=time.

I guess that means the rest of my working out is correct then?

Thanks Stuh 505

In most cases there will still be infinite solutions. Even for problems where it does narrow it down to 1 solution, solving is not straightforward. For physics problems, you will never be asked to solve a system of equations that has more unknowns than equations. It is easy to solve it using 2 equations like I showed you.

The work that you did leading up to this point looked right, but what I posted was a complete solution from start to finish, it does not start where you left off. Some of what you did is included in what I wrote.

In order to have "barely" a collision, the fast car would decellerate just enough so that by the time it was just about to hit the slow car, they were going the same speed. In this example we are assuming that the cars are "points" that have no size...so the ending condition is when the 2 points are at the same position and going the same speed. Thus, the value for "a" is the minimum value to have a collision. Your teacher wants the least amount of acceleration to avoid a collision, so that would be "a + epsiolon", where epsilon is some infinitely small number.

Notation:

A = fast car
B = slow car

f = final time
0 = start time

A.v(f) = final velocity of car A
B.v(f) = final velocity of car B
A.v(0) = initial velocity of car A...
A.x(0) = initial position of car A
A.x(f) = final position of car A...

Yes, but it will be an inequality right? and from there one can find the minimal value?

"But we know that for all real and positive values of "t" that
0.5at^2 + 58 2/3 t -600 <0
must be satisfied, therefore, from that, can we derive the values of "a" that will satisfy the inequality? The graph must open downwards (if you know what I mean), it must be completely below the x- axis, therefore a<o and the roots are non-real right?"

The roots of an equation

ax^2 + bx + c =0 is

x = (-b +- SQRT(b^2 - 4ac))/2a

Therefore if the roots of
0.5at^2 + 58 2/3 t -600

are non-real, then
b^2 - 4ac< 0

i.e.

(58.67)^2 - 4(0.5)(-600)<0

that's what I'm getting at.



Thanks, you helped me a lot! thanks a lot!

any one?

Isn't this a much simpler question? The first car is travelling relative to the second at 88 - 29 1/3 = 58.666666666 etc m/s and only has 600 metres to match velocity...

So simply sovle 58.66666^2 - 0^2 = 2 a * 600

a = ((58 2/3)^2/ 1,200)^1/2

a = 1.68 m/sec/sec deceleration to avoid a collision.

Once you realise you can work with the relative, rather than absolute velocities - the problem is trivial.


Oh, missed your train question,

You need equations v = u + at, s = ut + 1/2 at^2 and and v^2 - u^2 = 2as. Assume the train has length s, and find distance s1 so that the train moves only half of the time alloted t.

a = (v^2 - u^2)/2s, and t = (v-u)/a, so

t = 2s(v-u)/(v^2 - u^2), now ask in half this time say t1 how big is s1 as a proportion of all of s?

Note I realise the train is under constant acceleration - imply not only is a static, but far more helpfully there are no compression waves going through the carriages in the train - changes the distances between the end points you are considering - back in a sec...

so half this time is simply t1 = s(v-u)/(v^2 - u^2), so how far in terms of s1 will the train travel in this amount of time? (Note you don't have to calculate v1 as we are only interested in 1/2 of t, which we can express as factions of the end points.

s1 = ut1 + 1/2 a * t1 ^2 , so substitute in back a and t1 and simplify

s1 = u[s(v-u)/(v^2 - u^2)] + 1/2[(v^2 - u^2)/2s][s(v-u)/(v^2 - u^2)]^2

s1 = s [u(v-u)/(v^2 - u^2)] + s{1/2[(v^2 - u^2)/2][(v-u)/(v^2 - u^2)]^2}

S1 = s { [u(v-u)/(v^2 - u^2)] + 1/2[(v^2 - u^2)/2][(v-u)/(v^2 - u^2)]^2 }

so s1 / s percent is = 100 * [u(v-u)/(v^2 - u^2)] + 1/2[(v^2 - u^2)/2][(v-u)/(v^2 - u^2)]^2

... I'll leave you to simplify and check! Nifty being able to simply it only in terms of v and u!

Re: Equations of Motion (help)


I think the question should ask about the maximum acceleration, not the least acceleration

I think semantics aside (minimum deceleration, maximum acceleration) the questions are answered rather definitively!

Thanks a lot g_day for your help!

And yes, I think the question meant "what is the minimum deceleration", since isn't deceleration only negetaive acceleration.

But thanks a lot once again everyone.