Finding time given constant acceleration, initial velocity, & distance

  • Thread starter Brent. T
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  • #1
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Homework Statement



I want to cover a distance of 8,ooo,ooo,ooo,ooo m and I want to know the time it will take if I have an initial velocity of 176400 m/ sec with an acceleration of 19.6 m/sec2. I think I have the right equation here I'm just not sure how to solve for t. Could someone walk me through the steps? Cheers.

Homework Equations



D = 0.5* a * t2 + V0 * t

The Attempt at a Solution



8000000000000 m = 0.5* 19.6 m/sec2 * t2 + V0 * t

8000000000000 m = 9.8 m/sec2 * t2 + 176400 m/ sec * t
 

Answers and Replies

  • #2
Simon Bridge
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That is correct - the equation is quadratic in t ... so, put it in standard form and use the quadratic equation:
$$ax^2+bx+c=0 \implies x\in \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
 
  • #3
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Would the standard form be: 9.8t2 + 176400t + 8000000000000
 
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  • #4
haruspex
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Would the standard form be: 9.8t2 + 176400t + 8000000000000
Not quite. Watch the signs.
Also, sounds like this could get relativistic. Are you supposed to consider that?
 
  • #5
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9.8t2 + 176400t - 8000000000000 (?)

If it gets relativistic then I guess I have to consider it. This isn't for homework or anything like that. This is just for something I'm writing. I just wanted a quick solution to what I thought was a simple problem but I can see that it's escalating into a big problem. Even this quadratic stuff is over my head to be honest.
 
  • #6
Simon Bridge
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##(9.8)t^2 + (176400)t - (8000000000000)=0## .. these numbers look familiar.
Anyway: You posted this in the homework section - which produces a particular kind of response.

What education level are you at?
 
  • #7
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Sorry. The general section said that any homework/ course type questions should be posted here. Mine seemed to fit into that category.

I finished highschool a few years ago now but since I've never had reason to use anything above basic math I've forgotten most of what I learned regarding algebra.

I was really just looking for a way to calculate velocities. I did find a website to help me solve quadratic equations which, after putting in the variables from the standard form equation, has given me 894552.7270791921 and -912552.7270791921. I'm supposed to use the positive answer aren't I? So t = 894552 seconds. That sounds about right to me.
 
  • #8
haruspex
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So t = 894552 seconds.
At which point the velocity will be about 6% the speed of light, right? So maybe no need to bother Herr Prof Einstein.
 
  • #9
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Thanks for your help guys. Now that I know what to do with those velocity equations I think I should be set.

Cheers.
 
  • #10
Simon Bridge
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I'm supposed to use the positive answer aren't I? So t = 894552 seconds.
... indeed. The two answers are because there are two times the ship could pass through the destination position, at a constant acceleration, given it's present position. One of those times is in the past (the negative number) but you want the one in the future since you need to know how long to get there rather than how long since the ship was last there.

894552s is about 10days 8.5hours - well done.
 

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