Acceleration troubles

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Banksy
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Acceleration troubles

Postby Banksy » Tue Oct 07, 2008 8:13 pm UTC

Hey, residents of XKCD... I'm having a bit of bother with my Maths Mechanics homework, and was hoping someone here could help me.
I'll type out the question.
A skier leaves a lodge and travels on a horizontal line with constant velocity 7ms-1. Another skier, starting from rest, leaves one second later and tries to catch up with the first skier, by constantly accelerating at 1.5ms-2. Calculate the distance from the lodge that they meet.

Now I'm pretty sure (but I could be wrong) that to figure this out you need to involve a velocity time graph for the two skiers, and find the point where the distances are equal... and you do that by doing something involving equations to get them to be the same.
Any help is largely appreciated.
Thanks guys :)
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gorcee
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Re: Acceleration troubles

Postby gorcee » Tue Oct 07, 2008 8:24 pm UTC

So, a constant velocity means that the skier has zero acceleration.

That means his distance traveled over time is basically his speed times the time he skis. Just like if you're in a car doing 30 km/h for 2 hours, you go 60 km. If you know calculus, then this is the integral of the velocity.

Now acceleration is slightly trickier. Acceleration means that the skier is going faster and faster all the time. So the distance traveled is the integral of the velocity, which itself is the integral of the acceleration.

A way to look at the question is, "how FAST is the second skier going after T seconds?" Well, you handle this just like you handled the first problem. Multiply his acceleration (meters per second squared) by the time (seconds). Units cancel just like numbers. m/s^2 * s = m/s.

Then you have a profile of his velocity over time. So it starts off small, but gets bigger and bigger. This is OK. Just multiply the new relation for his velocity by time, and you have the distance!

This is only the first step in understanding. We don't KNOW how long or how far the 2 skiers travel before meeting. All we know is they meet somewhere. Let's call that distance D and the amount of time it takes T. Since both skiers are going D meters, set the two equations obtained equal to each other, and cancel out, and you should have an easy to solve quadratic equation.

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Banksy
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Re: Acceleration troubles

Postby Banksy » Tue Oct 07, 2008 8:37 pm UTC

I'm still quite lost here... I'm sorry :(
So should I be looking at something like
T x V = T x (T x a)
With the first skier on the left and the second on the right?
And both of those are equal to distance?
I'm confused :(
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btilly
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Re: Acceleration troubles

Postby btilly » Tue Oct 07, 2008 10:59 pm UTC

Banksy wrote:Now I'm pretty sure (but I could be wrong) that to figure this out you need to involve a velocity time graph for the two skiers, and find the point where the distances are equal... and you do that by doing something involving equations to get them to be the same.

That is your mistake. You don't want to write velocity as a function of time, write position as a function of time.

You will wind up with one equation per person. Then you just need to solve for what time the two equations give the same answer. Which should be a quadratic equation that you use the quadratic formula for.
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Alias
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Re: Acceleration troubles

Postby Alias » Tue Oct 07, 2008 11:19 pm UTC

use the equations of motion, and do them simultainiously

Spoiler:
arrange it so that skier one has an equation equal to distance
do the same for skier 2
set them equal to eachother
solve for distance


is how i would do it
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stormgren
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Re: Acceleration troubles

Postby stormgren » Wed Oct 08, 2008 1:40 am UTC

T x V = T x (T x a)


This is wrong... Let's take a step back: do you know any calculus? Failing that, do you know any constant acceleration equations (given this problem, you should)?

brodieboy255
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Re: Acceleration troubles

Postby brodieboy255 » Wed Oct 08, 2008 8:50 am UTC

It may be easier to let t=0 when the second skier starts, as you know how far the first skier is

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mochafairy
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Re: Acceleration troubles

Postby mochafairy » Wed Oct 08, 2008 9:34 am UTC

acceleration troubles? I say floor it! Oh, wait...

some basics:
  • the derivative of displacement with respect to time is velocity
  • the derivative of velocity with respect to time is acceleration
  • the derivative of acceleration with respect to time is called jerk

therefore:
  • the integral of jerk with respect to time is acceleration
  • the integral of acceleration with respect to time is velocity
  • the integral of velocity with respect to time is displacement

You have a velocity v. time function and an acceleration v. time function. You want to integrate both functions so they are displacement v. time functions. set them equal to each other and solve. (this is what Alias said to do, except more lengthy)
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Torn Apart By Dingos
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Re: Acceleration troubles

Postby Torn Apart By Dingos » Wed Oct 08, 2008 10:08 am UTC

Oh snap!

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Govalant
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Re: Acceleration troubles

Postby Govalant » Thu Oct 09, 2008 4:50 pm UTC

Torn Apart By Dingos wrote:Oh snap!


Fix'd!
Now these points of data make a beautiful line.

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