Calculate Constant Velocity & Speed for Trapezoid Move

[robertor]

Hi All,

A question to calculate the speed at constant velocity during a Trapezoid move.

An object moves from standstill at a known acceleration (1m/s/s) to a constant velocity (unknown) for a period (unknown) and then decelerates at a known deceleration (2m/s/s) to standstill. The total displacement is known (10m) and the total time is known (25s). How can I calculate the constant velocity and therefore the speed at constant velocity during the trapezoid move?

Thanks!
Rob

 

[tiny-tim]

welcome to pf!

Hi Rob! Welcome to PF!

Call the times t₁ t₂ and t₃

then write out all the equations …

show us what you get. ​

 

[robertor]

Which equation?

Thank you for the quick reply Tim!

I don't really know where to start due to the amount of unknowns...

If I begin with finding x for t₁:

x = v0t₁ + ½at₁²
x = 0*t₁ + ½*1*t₁²
x = ½*t₁²

Is this in the right direction for t₁?

 

[tiny-tim]

yes

now find the speed at t₁, which will be the starting speed for the next part,

and then write the equation for t₂

 

[robertor]

Constant Acceleration - Final Velocity

Final Velocity
v₂ = v₁ + at
v₂ = at

This gives me final velocity after the constant acceleration...but I am at a bit of a loss as to what to do here. Thanks for the help so far, you are very good!

 

[robertor]

Constant Velocity Displacement

Does this mean that the displacement during the constant velocity is calculated:
x₂ = (at)²

 

[tiny-tim]

(been out all evening! )

Final Velocity
v₂ = v₁ + at
v₂ = at

This gives me final velocity after the constant acceleration...but I am at a bit of a loss as to what to do here. Thanks for the help so far, you are very good!

(what is  ?)

yes v₁ = v₂ = t₁

t₂ is unknown

now work backwards from the end, to find the speed at the start of the deceleration, the same way you worked forwards from the start to find the speed at the end of the acceleration

 

[robertor]

Decel Part

OK, so working backwards for t3:

x = v0t₃ + ½at₃²
x = 0*t₃ + ½*2*t₃²
x = t₃²

The equation is simplified to this only because of the value that I passed in, had the decel not been 2m/s/s, the equation would look different, obviously!

 

[tiny-tim]

ok, now add all the x's, and all the t's

 

[robertor]

OK, so adding...

ACCEL: x = ½*t₁²
C.VEL: x = (at₂)²
DECEL: x = t₃²

x = ½*t₁² + (at₂)² + t₃²

 

[tiny-tim]

(just got up :zzz:)

C.VEL: x = (at₂)²

no, x = vt₂

(and then put the total displacement = 10, and write out total time = 20)

 

[robertor]

Simplifying

Thank you for the info so far!

x = (½*t₁²) + (vt₂) + (t₃²)
10 = (½*t₁²) + (vt₂) + (t₃²)

But I only have total time, and each of these time periods are separate, so how would I "put it" total time=20 into this equation?

 

[tiny-tim]

… how would I "put it" total time=20 into this equation?

you have to learn to translate from english to maths!

"total time" -> t₁+ t₂ + t₃ !​

(and now you'll need those two equations for v)

 

[robertor]

Sure, well I know:

t = t₁ + t₂ + t₃
x = 10

So,

10 = (½*t₁²) + (vt₂) + (t₃²)

But I am stuck here. If during static velocity (t₂) v=x₂/t₂ then v = tx₂. But this doesn't really help does it?

 

[tiny-tim]

you have t₁+ t₂ + t₃ = 25

and (½*t₁²) + (vt₂) + (t₃²) = 10

but you don't know what v is, sooo …

now you use your two equations for v, the one working forwards from the start, and the other working backwards from the finish! ​

(i'm going out now for the rest of the day)

 

[robertor]

Confused...

But I only need to find v, which is not part of the acceleration period or deceleration period, so why do the first two equations have anything to do with v (the constant velocity period)?

 

[tiny-tim]

you need v for the total displacement equation,

and you find v (as a function of t₁ or t₃) from the acceleration and deceleration equations

 

[robertor]

This equation?
v=v₀+at

 

[tiny-tim]

yes, use it forwards from the start, and backwards from the finish

 

[robertor]

Im really stuck...
I can't seem to figure out how v=v₀+at for the acceleration deceleration period will help me find v? Am i being stupid? I have been looking at this for hours and can't seem to slot the final pieces together.

t₁+ t₂ + t₃ = 25
(½*t₁²) + (vt₂) + (t₃²) = 10

How does using v=v₀+at help me from here?

 

[tiny-tim]

How does using v=v₀+at help me from here?

let's take the first stage …

that's v = 0 + 1*t₁, so you can substitute that into

(½*t₁²) + (vt₂) + (t₃²) = 10​

which was a four-unknowns equation (t₁ t₂ t₃ and v)

but now becomes a three-unknowns equation (t₁ t₂ and t₃ )

then you use v = v₀ + at again, for the last stage, and that gives you an equation relating v and t₃, so you can eliminate one more unkown

 

[robertor]

I see..

Oh I see, sorry i was confused as v was the constant velocity during t2, and I was substituting in the velocity as a result of acceleration, but you made me realize that this velocity is the final velocity of the acceleration period

v=v₀+at₁
v=at₁

So inserting this into the equation:

(½*t₁²) + (a*t₁*t₂) + (t₃²) = 10

For the decel:

v=v₀+at₃
v=2*t₃
t₃ = v/2
t₃ = (at₁)/2

So putting this into the equation:

(½*t₁²) + (a*t₁*t₂) + (((at₁)/2)²) = 10

(½*t₁²) + (a*t₁*t₂) + ((at₁)/4) = 10

 

[tiny-tim]

(½*t₁²) + (a*t₁*t₂) + (((at₁)/2)²) = 10

(½*t₁²) + (a*t₁*t₂) + ((at₁)/4) = 10

yes , except:

i] i think you meant ((a²t₁²)/4)

ii] you know that a = 1​

now simplify that other equation, t₁ + t₂ + t₃ = 25

 

[robertor]

Apologies, I typed it incorrectly.

(1/2*t₁²) + (1*t₁*t₂) + ((1²t₁²)/4) = 10

(1/2*t₁²) + (t₁t₂) + ((2t₁²)/4) = 10

(1/2*t₁²) + (t₁t₂) + ((t₁²)/2) = 10

(t₁²) + (t₁t₂) = 10

I still have 2 unknowns, but I also know t₁ + t₂ + t₃ = 25. I have got to a stand still again!

 

[tiny-tim]

hi robertor!

I still have 2 unknowns, but I also know t₁ + t₂ + t₃ = 25.

ok, now eliminate t₃ from that equation

 

[robertor]

how?

 

[tiny-tim]

same way you eliminated it from (½*t₁²) + (a²*t₁²*t₂) + (t₃²) = 10

(you are allowed to use things twice! )

 

[robertor]

Sorry, I read it over and I realize how dumb I sounded! I used this only a few steps ago!

So I have:
t₃ = (at₁)/2
t₃ = (2*t₁)/2
t₃ = t₁

t₁ + t₂ + t₃ = 25
t₁ + t₂ + t₁ = 25
(2t₁) + t₂ = 25

(t₁²) + (t₁t₂) = 10

 

[tiny-tim]

I used this only a few steps ago!

he he!

(2t₁) + t₂ = 25

(t₁²) + (t₁t₂) = 10

right, so t₂ = 25 - 2t₁,

and you can substitute that into the second equation

 

[robertor]

Simplifying Down...

(t₁²) + (t₁(25 - 2t₁) = 10

 

[tiny-tim]

ok, that's a quadratic equation in t₁

 

[robertor]

(t₁²) + (t₁(25 - 2t₁) = 10
t₁² + (25*t₁ - 2t₁²) = 10
25*t₁ - 1t₁² = 10
25*t₁ - 1t₁² - 10 = 0
-25*t₁ + t₁² + 10 = 0
-25*t₁ + t₁² = -10
-25*t₁ + t₁² = -10
-25*t₁ + t₁² + 156.25 = -10 + 156.25
-25*t₁ + t₁² + 156.25 = 146.25
(t₁ + -12.5)(t₁ + -12.5) = 146.25
t₁ + -12.5 = 12.093386622 OR -12.093386622
t₁ = 24.593386622
t₁ = 0.406613378

 

[tiny-tim]

t₁ = 24.593386622
t₁ = 0.406613378

yes, that looks ok: you have two possible solutions for t₁, so now find t₂ and t₃ in each case and check that it all works

(for something this long, there's very likely a mistake somewhere! )

finally, the original question was …

A question to calculate the speed at constant velocity during a Trapezoid move.

… so find v

btw …

(t₁²) + (t₁(25 - 2t₁) = 10
t₁² + (25*t₁ - 2t₁²) = 10
25*t₁ - 1t₁² = 10
25*t₁ - 1t₁² - 10 = 0
-25*t₁ + t₁² + 10 = 0
-25*t₁ + t₁² = -10
-25*t₁ + t₁² = -10
-25*t₁ + t₁² + 156.25 = -10 + 156.25
-25*t₁ + t₁² + 156.25 = 146.25
(t₁ + -12.5)(t₁ + -12.5) = 146.25
t₁ + -12.5 = 12.093386622 OR -12.093386622

most people would just have written

(t₁²) + (t₁(25 - 2t₁)) = 10
t₁² - 25*t₁ + 10 = 0
so t₁ = 12.5 ± √(156.25 - 10)
= 24.59 or 0.41 …​

… is there any reason why you've avoided that?

 

[robertor]

Finally...

Solution A:
t₁ = 24.593386622
v = v₀ + at
v = 1 * 24.593386622
v = 24.593386622m/s

Solution B:
t₁ = 0.406613378
v = v₀ + at
v = 1 * 0.406613378
v = 0.406613378m/s

… is there any reason why you've avoided that?

I found it easier to do the steps logically:
t1*25 - 2t₁²

And then work out the steps one by one, moving the constant to the left hand side, so the result is 0. Multiplying by -1 to remove the negative figures. Half of part a, squared added to both sides. Thats the way I was taught in I think? Your method seems more concise however I fail to see the middle steps you took, possible because I am nowhere near as advanced as you are!

 

[tiny-tim]

Solution A:
t₁ = 24.593386622
v = v₀ + at
v = 1 * 24.593386622
v = 24.593386622m/s

Solution B:
t₁ = 0.406613378
v = v₀ + at
v = 1 * 0.406613378
v = 0.406613378m/s

yes that looks fine …

now round it off to a sensible number of significant figures, and you're finished!

(and, as i said before, there's probably a mistake somewhere, so you'd better check it very carefully)

Thats the way I was taught in I think? Your method seems more concise however I fail to see the middle steps you took …

i used the standard quadratic equation formula …

first convert the equation to ax² + bx + c = 0, and then the solution is x = (-b ± √(b² - 4ac))/2 …

it's completely acceptable in exams, so you should learn it! ​

ok, lets' go over how you did it, so you know what to do in future …

first, you translated some of the question from maths into english … that gave you two equations

second, you used physics to get two more equations, making 4 equations with 4 unknowns

third, you started eliminating one unknown at a time until you got to 1 equation with only 1 unknown