# Linear acceleration with letters instead of numbers

1. Feb 16, 2013

### Woolyabyss

1. The problem statement, all variables and given/known data
A lift starts from rest and travels with constant acceleration 4 m/s^2.It then travels with uniform speed and comes to rest with constant retardation of 4 m/s^"
The total distance traveled is d and the total time taken is t.

Show that the time spent travelling at a constant speed is (t^2 - d)^(1/2)

2. Relevant equations
I called the maximum velocity "v", the time spent accelerating "t1", the time spent at constant speed "t2" and the time spent decelerating "t3" distance traveled when accelerating "d1",when at constant velocity "d2" and when decelerating "d3"

3. The attempt at a solution
I used v = u + at, to find that t1 = v/4 and t3 = v/4.
I then multiplied the times by (1/2)v
to get the distances in both cases d1 = v^2/8 and d3 = v^2/8.
I then said d2 = vt2 =d - (v^2)/4
I divided v by both sides to get t2= (d/v-(v/4).
I made t1 + t2 + t3 = t and simplified to get (4d + v^2) / (4v) = t.
I tried to isolate v but couldn't since there was a v squared.
My main problem here I think is I cant get v on its own so i can't get rid of v in t2.
Any help would be appreciated.

2. Feb 16, 2013

### tms

What you are looking for, in your notation, is t2 in terms of t and d. Don't worry about v. Write down the kinematics equations for each part of the path, using forms that only involve distance, time, and acceleration, since that is what you are given.

3. Feb 16, 2013

### Woolyabyss

Am I not required to to use the maximum speed "v" since d2 has no acceleration.The only way I can express it is in terms of speed and time isn't it?

4. Feb 16, 2013

### tms

You can do that if you want, but it would be simpler in the end to express that speed in terms of what you already know.

5. Feb 17, 2013

### xodin

Hmmm, after looking at this problem, does it not require that his t1=t3, or am I wrong? I get the right answer if I assume t1=t3 I think.

6. Feb 17, 2013

### haruspex

The deceleration during t3 is equal and opposite to the acceleration during t1. The initial and final velocities are the same. Doesn't that imply t1=t3?

7. Feb 17, 2013

### xodin

Yep, when I first read it though I didn't think it specified that it started and stopped at rest.