1D Kinematics

1. Sep 23, 2013

fogvajarash

1. The problem statement, all variables and given/known data
Captain Kirk is on an important space mission to bring a Klingon ambassador to space outpost 13. He must travel 6000. km in less than 30.0 min, or else war will be declared between the Klingons and humans. After traveling exactly 27.0 min at a steady pace, he sees that there are still 870.0 km left to go.
His ship must then accelerate at 20.000 m/s-2 for how many seconds in order to achieve the desired time?

2. Relevant equations
s = u + at
s = vot + 1/2at2
v2 = u2 + 2as

3. The attempt at a solution
I've tried separating the motion of the spacecraft in 3 different times. Time 1 is when the spacecraft is at constant speed until 27.0min, time 2 is when the spacecraft starts to accelerate at 20.000m/s-2 and time 3 is when the spacecraft travels at the new constant speed increased by the acceleration. So:

For time 1:

s = vot

Here i found the speed just by plugging in s = 5130000m and t = 1620s.

For time 2 and 3:

s = vot + 1/2at2 + vf(180-t)

180-t as the spacecraft has 3 minutes remaining (180s). I found the vf with the following equations:

vf = v0 + at

I've tried this problem numerous times and I keep getting that the result is 0.044s, but it is wrong.

2. Sep 23, 2013

HallsofIvy

Staff Emeritus
You are not taking into account the distance traveled while accelerating.
Yes, he must go 870 km in 3 m (870000 m in 180 s). If he accelerates at 20 m/s^2 for T seconds. he will have gone a distance $10T^2+ 3166.67 T$ and will have achieved a speed of 20T m/s. He must have $$10T^2+ 3166.67T+ (20T)(180- T)= 870$$.

(Knowing Kirk, he will probably dawdle in order to prolong the war!)

3. Sep 23, 2013

Staff: Mentor

@HallsofIvy: He did that into account (at least in the equation s=...). I think you are mixing meters and kilometers here.

@fogvajarash: Here is a quick way to see that 0.044s cannot be right:
With an acceleration of 20m/s^2, accelerating for 0.044s changes the speed just by roughly 1m/s, from 3167 to 3168m/s. To travel the remaining 870km in 3 minutes, the average speed (which is lower than the maximal speed) has to be of the order of 4800m/s. The acceleration time has to be much more than 0.044s.

Please show your work. Your approach looks right, so we have no way to tell what you did wrong if you don't show it.

By the way: m/s-2 does not make sense. ms-2 = m/s2 are correct units for an acceleration.

4. Sep 23, 2013

fogvajarash

Crap I meant 20000m/s2 instead of 20.0. Sorry for the mistake

5. Sep 23, 2013

Staff: Mentor

That's still not enough, based on the same estimate.

And that acceleration will crush every living object on the ship.

6. Sep 23, 2013

fogvajarash

I suppose the problem isn't that realistic haha.

Here's my procedure:

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7. Sep 23, 2013

Staff: Mentor

There are 870km left, not 730km.

8. Sep 23, 2013

fogvajarash

I get as a final result t = 0.0834s. Is this result accurate? (I've changed the 730000 by a 870000).

9. Sep 23, 2013

Staff: Mentor

You can use this value and plug it in your initial equation for s= to check it. It looks reasonable, but I did not check it in detail (WolframAlpha can do that better anyway).

10. Sep 23, 2013

fogvajarash

I keep getting an incorrect result for the exercise. Have i made a mistake with setting up the equations? It looks alright to me.

11. Sep 23, 2013

Staff: Mentor

The equations look good.
Are you sure the acceleration is 20000m/s^2? That number looks really odd compared to the other numbers, 20m/s^2 would give more reasonable results.