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
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.
Now we know two other pieces of information:
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:
. (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.)
So we are looking for t2.
Well, solve the t equation for t1:
and put this value of t1 into the d equation: