Proving Distance Traveled With Formula

In summary, 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.
  • #1
JDK
27
0
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
 
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  • #2
Eq u can use are 0=vo-at
d=vot-1/2at^2
 
  • #3
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
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 textbook 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 textbook 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
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?

They are just subscripts



I recall in my textbook 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?

rectangular area will represent a acceleration = 0
Whereas Triangular area do represent accelerated/decelerated motion depending upon the slope of the equation
 
  • #6
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...

|
|
|
|
|
|
|-----------------------
|
|
|
|_______________________

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
hmmm u basically needs some kinematics equations for constant acceleration which i believe u know
d = (vo)²/a - 1/2(a)(ts)²
is a littly fuzzy u can get to it via the equations its just rearrangements

Graphically, u would get[tex]\frac{v_0}{2}*t[/tex] OR [tex]\frac{at^2}{2}[/tex]
 

What is the formula for calculating distance traveled?

The formula for calculating distance traveled is distance = speed x time. This formula assumes that the speed is constant throughout the entire journey.

Can this formula be used for any type of motion?

Yes, this formula can be used for any type of motion as long as the speed remains constant. This includes motion in a straight line, circular motion, and even motion in multiple directions.

What are the units for distance, speed, and time in this formula?

The units for distance can vary depending on the system of measurement used, but common units include meters, kilometers, miles, etc. Speed is typically measured in meters per second, kilometers per hour, miles per hour, etc. Time is usually measured in seconds, minutes, or hours.

Can this formula account for changes in speed during the journey?

No, this formula only works if the speed remains constant. If there are changes in speed, a more complex formula, such as the displacement formula, must be used to accurately calculate distance traveled.

Are there any limitations to this formula?

Yes, this formula only works if the speed remains constant. It also assumes that the motion is occurring in a straight line. If the motion is not in a straight line or if the speed is changing, this formula will not accurately calculate the distance traveled.

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