Calculate Constant Velocity & Speed for Trapezoid Move

[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

 

[robertor]

Rounding

v = 24.59m/s OR v = 0.41m/s

I mean looking at it, I know which is more likely, is there a way to confirm which of these two values is correct?

 

[tiny-tim]

… looking at it, I know which is more likely, is there a way to confirm which of these two values is correct?

yup!

something comes out negative when it can't be

in most problems, you'd find that t₁ has one positive solution and one negative solution …

obviously, you reject the negative solution!​

but in this problem, it's slightly different …

can you see what it is? ​

 

[robertor]

I could calculate the displacement during t1, t2 and t3. They should add up to my known displacement, if they don't, then its the other one?

 

[tiny-tim]

no

what is the value of t₂ in each case? ​

 

[robertor]

Sure I can calculate the value of t₂ in this scenario, however my next task is to rewrite this equation as a generic formula to find t₁, using the same technique you have shown me. Therefore, sure I can look at t₂ in both cases but how will I know which one is correct. If I provide accel, decel, total time and total displacement, what happens if the object does not have time to accelerate to a constant velocity, and the shape within the velocity x time graph becomes a triangle? Surely t₂ could be negative using this current equation. There must be a more generic method for checking which value of V is correct, which is why I suggested adding up the displacements or the times to see if they equal the displacement or time provided at the beginning :)

 

[tiny-tim]

… which is why I suggested adding up the displacements or the times to see if they equal the displacement or time provided at the beginning :)

that can't possibly work: the correct values for total time and displacement were part of your input

Therefore, sure I can look at t₂ in both cases but how will I know which one is correct.

calculate both values, and then it should be obvious

what are they?​

 

[robertor]

Sure on this occasion the correct answer is obvious. However what I was pointing out is that the answer will not always be obvious, am I wrong?

 

[tiny-tim]

Sure on this occasion the correct answer is obvious. However what I was pointing out is that the answer will not always be obvious, am I wrong?

i'm not entirely convinced that you have got the correct answer, since you haven't specifically said what it is

the general rule is that if there's two solutions, then they're both valid …

why would they not be??​

the only exception is where something is undefined for particular values (in this case for negative values) of some variable

 

[robertor]

t₂

t₂ = 24.390079932 OR t₂ = -11.890079933

 

[tiny-tim]

yup! and, as i think you've noticed, t₂ can't be negative (unless you redefine the question considerably), so the second solution has to be rejected

(but plenty of problems do have two solutions, eg if you throw the ball at 5 m/s, at what angle should you throw it to get it through the hoop?)

 

[robertor]

What if both answers are positive?

 

[tiny-tim]

then both answers are valid