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.

Hello, anyone who can help, please please do....

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)

Please help solve!!!!!
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: