Show that s = [fr/(f + r)]t^2/2 where f & r are accel and decel

[physicsnnewbie]

Homework Statement

A train travels dist 's' in 't' secs. It starts and ends at rest. In first part of journey it moves with const accel 'f', and in second part with const decel 'r'. Show that s = [fr/(f + r)]t^2/2 where f & r are accel and decel

Homework Equations

The Attempt at a Solution

I spent at least an hour trying to figure it out, but not sure how

 

[tiny-tim]

hi physicsnnewbie!

i had to read the question three times before it dawned on me that the vital part is …

It starts and ends at rest.

ok … what is your equation for this piece of information?

 

[hunt_mat]

The other vita part of the information is there nowhere where it moves at constant velocity, so the only things to have to calculate, the distance in acceleration and deceleration.

Suppose the maximum speed the train is v, then...

 

[physicsnnewbie]

v = ft (velocity section 1)
v = v0 + rt (velocity section 2)

How do i find the initial velocity v0 for section 2

 

[hunt_mat]

Read my post, there was no point at which the train was traveling at constant speed.
What you should have is two times, v=ft_{1} and v=-rt_{2}, where t_{1}+t_{2}=t. Do the same for s and then relate s and t together. Lots of algebra.

 

[SammyS]

v = ft (velocity section 1)
v = v0 + rt (velocity section 2)

How do i find the initial velocity v0 for section 2

As mentioned elsewhere, use t₁ for the time of acceleration. Then t-t₁ is the time interval over which deceleration occurs.

The velocity at the t₁ is v = f*t₁. (This is the maximum velocity, also it's the velocity at the start of the deceleration.) So to answer your question, v₀ for the deceleration portion is v.

So 0 = v - r*(t-t₁). Use this equation along with v = f*t₁ to solve for t₁.

The other thing which will give you the desired answer is that s = v_avg*t, where v_avg is the average velocity for the entire trip.

Since you have constant acceleration for each leg of the trip, v_avg = (v + 0)/2. It's the same for each portion of the trip.

Put all that together, & you will be fine.

 

[physicsnnewbie]

Ok,

v₁ = ft₁
v₂ = ft₁+ r(t - t₁)

s₁ = (ft₁²)/2
s₂ = (ft²)/2 + r(t-t₁)²/2
s₁ + s₂ = (ft²)/2 + (ft²)/2 + r(t-t₁)²/2
= (2ft² + rt² -2rt₁t + rt₁²)/2

t₁ = -rt/(f-r)

So I substitute -rt/(f-r) for t₁ into the combined distance equation and with the help of an algebra solver I get:

ft²(2f² - 3fr + 2r²)/2(f - r)² which isn't correct. So I must still be doing something wrong. I didn't quite understand your average velocity equation at the bottom sammy.

 

[tiny-tim]

hi physicsnnewbie!

s₁ = (ft₁²)/2
s₂ = (ft²)/2 + r(t-t₁)²/2

nooo … your constant acceleration equation for s₂ should start with either initial speed or final speed times time

 

[accatagliato]

Hi, I'm italian student, excuse me for my english

The problem is the integration of the two differential equations, in fact:

s₁ = ft₀²/2

s₂ = ft₀(t-t₀) - r(t-t₀)²/2

in this case I put "r" positive.
After substitutions and calculations you'll arrive at the solution

 

[SammyS]

Ok,

v₁ = ft₁
v₂ = ft₁+ r(t - t₁)

s₁ = (ft₁²)/2
s₂ = (ft²)/2 + r(t-t₁)²/2
s₁ + s₂ = (ft²)/2 + (ft²)/2 + r(t-t₁)²/2
= (2ft² + rt² -2rt₁t + rt₁²)/2

t₁ = -rt/(f-r)

So I substitute -rt/(f-r) for t₁ into the combined distance equation and with the help of an algebra solver I get:

ft²(2f² - 3fr + 2r²)/2(f - r)² which isn't correct. So I must still be doing something wrong. I didn't quite understand your average velocity equation at the bottom SammyS.

As the train accelerates (constant acceleration of f ) to a final velocity, v, the average velocity is (v + 0)/2 because the initial velocity is zero, the final is v, and the acceleration is constant). So the average velocity is v/2, where v is the velocity at the end of the acceleration phase of the trip. (What you call v₁, I call v.)

Similarly, for the deceleration portion of the trip (constant acceleration of -r ), the train starts with velocity v, and ends with velocity 0, so v_avg = (v/2) for the portion also. The average velocity is the same for both parts of the trip, so for the entire trip, v_avg = v/2.

The thing about using average velocity is that it is defined as the displacement divided by the time. So, for the f leg, v_avg = s₁/t₁, for the r leg, v_avg = (s-s₁)/(t-t₁), and for the entire trip, v_avg = s/t . Therefore, s = (v_avg)·(t). (You can calculate s without having to square (t-t₁).

I'm sure that you should treat the deceleration, r, as a positive number, that is to say, for the deceleration leg of the trip, a = -r. Using this, I get t₁ = rt/(f+r).

Your first two equations above are:

v₁ = ft₁
v₂ = ft₁+ r(t - t₁)

Note: that v₂ = 0, and you should use -r for deceleration.

These then become:
v₁ = ft₁
0 = ft₁ - r(t - t₁) ... Solve this for t₁ and plug that into the equation immediately above for v₁.

Then s = (v_avg)·(t) = (v₁/2)·(t).

See how that works.

 

[physicsnnewbie]

Thanks for the help sammy and everyone else, I finally got there (with the answer pretty much spelled out) :). I can think about as laterally as a brick. I also solved it with the revised equation mentioned by accatagliato. Man that is a LOT of work. I'm surprised I actually solved it without obscuring the answer with errors! I did make one time consuing error though which i managed to spot.

One more thing. How do i derive the average velocity formula? I tried a couple times with no success.

Tiny-tim, could you elaborate on your last post.

 

[tiny-tim]

hi physicsnnewbie!

s₂ = 0 + 1/2 r(t - t₁)²

or s₂ = v_max(t - t₁) - 1/2 r(t - t₁)²

(btw, you should be able to do the whole problem in about 4 lines, so fiddle about with your answer and try to cut out any unnecessary details )

 

[SammyS]

Thanks for the help sammy and everyone else, I finally got there (with the answer pretty much spelled out) :). I can think about as laterally as a brick. I also solved it with the revised equation mentioned by accatagliato. Man that is a LOT of work. I'm surprised I actually solved it without obscuring the answer with errors! I did make one time consuming error though which i managed to spot.

One more thing. How do i derive the average velocity formula? I tried a couple times with no success.

Tiny-tim, could you elaborate on your last post.

Average Velocity: Use this method when acceleration is constant. Start with the easier case in which the initial velocity, v_i = 0, and the initial time is t = 0.

At time t, v = v_i + a·t  →  v = 0 + a·t  →  v = a·t. Look at the graph of velocity versus time. Velocity, v, will be a straight line segment with a slope of a. Since the velocity graph is a straight line, v_avg for any time interval is equal to the velocity at the midpoint of the time interval.

At some final time t = t_f, draw a vertical line segment up to the velocity graph. The velocity is v_f at time t_f and is given by, v_f = a·t_f. The area of the triangle formed by this vertical line, the time axis, and the velocity graph is Area = (1/2)·base·altitude = (1/2)·t_f ·v_f. Substituting, a·t_f in for v_f, gives: Area = (1/2)·t_f ·a·t_f  →  Area = (1/2)·a·(t_f)².

Of course, as it turns out, s, the distance traveled during time, 0 to t_f, is s = (1/2)·a·(t_f)². This is equal to the area under the velocity versus time graph, as you have probably been told.

One definition of average velocity is: v_avg = (displacement)/(time). Since we are dealing with one dimensional motion, and velocity is non-negative, displacement is the same as distance traveled, s. Therefore, between times t = 0 and t = t_f is v_avg = s/t_f.
So that, s = v_avg·t_f. This is true even if acceleration is not constant. However, when acceleration is constant, then v_avg = (v_i + v_f)/2 . Therefore, s = (1/2)·(v_i + v_f)·t_f. If v_i = 0, then s = (1/2)·v_f ·t_f.
           

 

[physicsnnewbie]

Thanks tim and sammy. By the way, the book i am using is Calculus: An intuitive and physical approach by Morris Kline. So I didn't know a lot of these formulas such as the average velocity formula (v₀ + v_f)/2 or when to use certain formulas. I think he probably intends for people to solve them intuitively without resorting to formulas which they don't understand. It is actually really well written and easy to understand, but some of the problems at the end of chapters seem to require concepts not discussed in the chapter, maybe hoping you can discover some of these concepts yourself. I don't mind this, except that I'm not very good at it.