# Proving Distance Traveled With Formula

1. Jan 8, 2004

### JDK

Hello,

I have a small question. I just don't know where to go with it. If anyone could tell me where to start at that would be awesome. Please and thank-you.

An automobile is braked to a stop with a uniform deceleration in a time of ts. Show that the distance traveled during this time is given by d = (vo)²/a - 1/2(a)(ts)²

2. Jan 9, 2004

### himanshu121

Eq u can use are 0=vo-at
d=vot-1/2at^2

3. Jan 9, 2004

### HallsofIvy

Staff Emeritus
The problem is that we don't know what tools you have to work with.

If you know that v= v0+ at and that d= (a/2)t2+ v0t+ d0 then himanshu121's suggest is exactly what you need.

You can derive both of those from the simple statement that acceleration is a constant but you need to use calculus.

4. Jan 9, 2004

### JDK

Tools? Well, I'm doing Physics 20 and am doing the unit on acceleration, and displacement during constant acceleration. Calculus is a tad bit out of my league ;).

Do the '0's in both of your equations have a significant meaning? All I know is that in my problem, the velocity (v) has a subscript '0', which is shown in my first post. I haven't seen that before. Is it just a subscript meant for distinguishing it from other (v) variables or does it hold meaning?

Thanks for bearing with me so far...

I recall in my text book that it mentioned (briefly) something about one part of an equation on a velocity time graph, with constant acceleration (as it is assumed throughout the Unit), representing the rectangular area under the curve, and the other part of the equation representing the triangular area under the curver. Thus, giving total displacement. Would my problem have to do with this?

Can't say I like my text book very much. Some of the questions I get from my teacher aren't covered in the book to a degree which allows me to have a clue. They're more - here's the gun - go shoot in the dark for a while and maybe you'll hit the target. But, I understand his reasons - forces students to learn to actually know 'why' things happen and how to apply that to different situations.

Anyways, thanks again!

5. Jan 9, 2004

### himanshu121

They are just subscripts

rectangular area will represent a acceleration = 0
Whereas Triangular area do represent accelerated/decelerated motion depending upon the slope of the equation

6. Jan 9, 2004

### JDK

Well, it's good to know '0' just means '0' and nothing more. I though things were about to get more confusing!

I understand what you mean with the rectangles and triangles. The graph I saw in my text which depicted what I was saying looked something like this...

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Imagine a line inclining upwards from the top left corner of the rectangular area that would eventually meet the right side of the graph to form a triangle after drawing a line straight down to the right side of the rectangular area.

I've been looking at the equations posted and I know both of them generally...

vf = vi + at
d = vi(t) + 1/2(a)(t)²

Proving the statement made in my original problem is still a bit fuzzy. What does one need to know, to know how to do this question? I feel the urge to pull my hair out violently.

*sits thinking*

7. Jan 9, 2004

### himanshu121

hmmm u basically needs some kinematics equations for constant acceleration which i believe u know
is a littly fuzzy u can get to it via the equations its just rearrangements

Graphically, u would get$$\frac{v_0}{2}*t$$ OR $$\frac{at^2}{2}$$