# Math Help - Confusing ALM(Accelerated linear motion)

1. ## Confusing ALM(Accelerated linear motion)

Hi guys, i saw this problem somewhere and thought it was pretty easy, but it driving me absolutely mental!Can any one help me solve it?

Here it goes.

A lift starts from rest with constant acceleration 4m/s. It then travels with uniform speed and finally comes to rest with constant retardation 4m/s, The total time taken is t and the total distance travelled is d. show that the time for which it travelled at uniform speed is
square root of ((t)squared - d)

2. Originally Posted by heebbyjeeby
Hi guys, i saw this problem somewhere and thought it was pretty easy, but it driving me absolutely mental!Can any one help me solve it?

Here it goes.

A lift starts from rest with constant acceleration 4m/s. It then travels with uniform speed and finally comes to rest with constant retardation 4m/s, The total time taken is t and the total distance travelled is d. show that the time for which it travelled at uniform speed is
square root of ((t)squared - d)

For starters, check your units. Acceleration is measured in " $m/s^2$" not m/s.

Also, not only is your "correct" answer incorrect, it makes no sense as a basic unit analysis will show. Your given equation says:
$t_2 = \sqrt{t^2 - d}$
Unitwise this says:
$s~=~\sqrt{s^2 - m}$

The units on both sides are obviously not equal.

Let's take this in stages. I'm going to set an origin where the lift starts, and use a +y direction straight upward. I'm going to take the speed of the lift when traveling at a constant velocity to be V.

So at the start of the motion we have that the lift starts from rest and accelerates uniformly to a speed V in a time t1. So
$v = v_0 + at$

$V = at_1$

So it was accelerating for a time
$t_1 = \frac{V}{a}$
where $a = 4~m/s^2$.

How far (y1) did it go during this time?
$y = y_0 + v_0t + \frac{1}{2}at^2$

$y_1 = \frac{1}{2}at_1^2$
since the lift started from rest and at the origin.

Now, let's talk about the constant velocity section. It will cover a distance y2 in a time t2:
$y = y_0 + v_0t + \frac{1}{2}at^2$

$y_2 = Vt_2$

Finally, let's talk about the "retardation" section. The lift has an acceleration of -a over this period and starts with a speed V. It slows to rest in a time t3. Similar to the first calculations we have that
$t_3 = \frac{V}{a}$ <-- $a = 4~m/s^2$ again. I already took care of the negative sign.
and
$y_3 = Vt_3 - \frac{1}{2}at_3^2$

Now we know two other pieces of information:
$d = y_1 + y_2 + y_3 = \frac{1}{2}at_1^2 + Vt_2 + Vt_3 - \frac{1}{2}at_3^2$
and
$t = t_1 + t_2 + t_3$

Now, think about the following for a minute. The lift starts from rest and accelerates uniformly to a speed V. At the end the lift is moving at a speed V and decelerates uniformly, with the same magnitude of acceleration as in the first part, to rest. I urge you to prove the following facts: $y_3 = y_1$ and $t_3 = t_1$. (You should be able to see this as a logical consequence of the problem, but it is good practice to use the motion equations to do it.)
$d = y_1 + y_2 + y_3 = \frac{1}{2}at_1^2 + Vt_2 + Vt_1 - \frac{1}{2}at_1^2 = V(t_1 + t_2)$
and
$t = 2t_1 + t_2$

So we are looking for t2.

Well, solve the t equation for t1:
$t_1 = \frac{t - t_2}{2}$
and put this value of t1 into the d equation:
$d = V \left ( \frac{t - t_2}{2} + t_2 \right )$

Now solve for t2:
$t_2 = \frac{2d}{V} - t$

This equation is radically different from the equation that you gave as the answer.

-Dan

Hey, thanks for replying, but the answer we are trying to prove is correct. It appeared in the leaving cert in Ireland. Also i think you are making the question to complicated, similar questions can be answered using a much lower knowledge of applied maths.

4. Originally Posted by heebbyjeeby
Hey, thanks for replying, but the answer we are trying to prove is correct. It appeared in the leaving cert in Ireland. Also i think you are making the question to complicated, similar questions can be answered using a much lower knowledge of applied maths.
Well, you see there are two problems:
1) I could "dumb it down" a bit (as you said in your PM), but I can't think of a way to do it much. Physics equations aside we are still going to be left with a system of two unknowns in two equations. I can't solve it by any easier method than I did. As for the rest of it, likely the problem is that you haven't seen how I define the variables:
$y_0$ is a y value at the initial (t = 0 s) time.
y is the y value at time t.

Similarly
$v_0$ is the value of the velocity at the initial time.
v is the v value at time t

The $t_1, t_2, t_3$ and $y_1, y_2, y_3$ represent the times and distances over each successive stage of the problem.

I am sorry that I can't rewrite this for you. That doesn't say much for me as a teacher, does it?

2) There is no way that your given answer can be correct. As I said, the units come out all wrong. So either someone has made a bunch of scaling definitions (to remove the constants), or someone has made a huge mistake. Or there is something about the wording of the problem I didn't understand, or was left out of the original.

I checked my answer before I posted it and I just rechecked it now. Given the problem as it was written I stand by my answer. You are going to have to seek another opinion if you are certain that your given solution is correct.

-Dan

5. ## You are both right

topsquark, the problem is that you haven't finished the question. There is no reason to believe that your answer is different to the one required.

What is really being asked for here is $t_2$ be expressed in terms of t and d. The other information in the result you are proving is only useful to check your work.

The following part of topsquark's clear and understandable working is all correct:

Quote:
Originally Posted by heebbyjeeby
Hi guys, i saw this problem somewhere and thought it was pretty easy, but it driving me absolutely mental!Can any one help me solve it?

Here it goes.

A lift starts from rest with constant acceleration 4m/s. It then travels with uniform speed and finally comes to rest with constant retardation 4m/s, The total time taken is t and the total distance travelled is d. show that the time for which it travelled at uniform speed is
square root of ((t)squared - d)

For starters, check your units. Acceleration is measured in "" not m/s.

Also, not only is your "correct" answer incorrect, it makes no sense as a basic unit analysis will show. Your given equation says:

Unitwise this says:

The units on both sides are obviously not equal.

Let's take this in stages. I'm going to set an origin where the lift starts, and use a +y direction straight upward. I'm going to take the speed of the lift when traveling at a constant velocity to be V.

So at the start of the motion we have that the lift starts from rest and accelerates uniformly to a speed V in a time t1. So

So it was accelerating for a time

where .

How far (y1) did it go during this time?

since the lift started from rest and at the origin.

Now, let's talk about the constant velocity section. It will cover a distance y2 in a time t2:

Finally, let's talk about the "retardation" section. The lift has an acceleration of -a over this period and starts with a speed V. It slows to rest in a time t3. Similar to the first calculations we have that
<-- again. I already took care of the negative sign.
and

Now we know two other pieces of information:

and

Now, think about the following for a minute. The lift starts from rest and accelerates uniformly to a speed V. At the end the lift is moving at a speed V and decelerates uniformly, with the same magnitude of acceleration as in the first part, to rest. I urge you to prove the following facts: and . (You should be able to see this as a logical consequence of the problem, but it is good practice to use the motion equations to do it.)

and

So we are looking for t2.

Well, solve the t equation for t1:

and put this value of t1 into the d equation:
now using $V = at_1$,
$V = 4t_1$
and

so
$V = 2(t-t_2)$
and

$d = 2(t-t_2) ((t-t_2)/2+t_2)$

now we solve for $t_2$:

$d = t^2-2tt_2 + t_2^2 + 2tt_2-2t_2^2$

$d = -t_2^2 + t^2$

$t_2^2 = t^2-d$

$t_2 = \sqrt{t^2 - d}$

By the way, unit analysis is not useful when a parameter is defined with a constant, in this case a = 4 $ms^-2$, because you cannot tell where the constant number of $ms^-2$ has been to make things even up.

topsquark, the problem is that you haven't finished the question. There is no reason to believe that your answer is different to the one required.

What is really being asked for here is $t_2$ be expressed in terms of t and d. The other information in the result you are proving is only useful to check your work.

The following part of topsquark's clear and understandable working is all correct:

now using $V = at_1$,
$V = 4t_1$
and

so
$V = 2(t-t_2)$
and

$d = 2(t-t_2) ((t-t_2)/2+t_2)$

now we solve for $t_2$:

$d = t^2-2tt_2 + t_2^2 + 2tt_2-2t_2^2$

$d = -t_2^2 + t^2$

$t_2^2 = t^2-d$

$t_2 = \sqrt{t^2 - d}$

By the way, unit analysis is not useful when a parameter is defined with a constant, in this case a = 4 $ms^-2$, because you cannot tell where the constant number of $ms^-2$ has been to make things even up.
I see what you are saying: for this particular value of the acceleration both of our expressions end up being the same. I didn't see that and I thank you for pointing it out.

I still stand firm, however that $\sqrt{t^2 - d}$ is an invalid expression. Perhaps I'm being anal about it (not the first time I've been accused of that) but when reporting a final result the units on a physical equation need to be specified, else that expression is meaningless. In a Mathematical equation we need not be so picky.

-Dan

7. Sorry, I misunderstood your previous post. You are right, there do need to be units in there.