Physics Question

[MrCodeDude]

The Starship Enterprise returns from warp drive to ordinary space with a forward speed of 50km/s. To the crew's great surprise, a Klingon ship is 100km directly ahead, traveling in the same direction at a mere 20 km/s. Without evasive action, Enterprise will hit the Klingon ship in a bit over 3.0s. The enterprises computers react instantly to brake the ship. What accelleration does the Enterprise need to barely avoid a collision with the Klingon ship? Assume accelleration is constant.

I know this:
Klingon Ship
Velocity Initial = Velocity Final = 20km/s
Accelleration = 0
Position Initial = 100

USS Enterprise
Velocity Initial = 50km/s ; Velocity Final = 0km/s
Accelleration = ?
Position Intial = 0

So, I need to find the accelleration.

Klingon equation for distance with respect to time is:
a(t) = 0
v(t) = 20
s(t) = 20t + 100

USS Enterprise equation is:
Sfinal = Sinitial + Vinitial * time + 1/2 acceleration * time^2
Sfinal = 0 + 50 * t + 1/2 * a * t^2
Sfinal = 50t + 1/2 * a * t^2

Vfinal = Vinitial + a*t
0 = 50 + a*t
-50 = a*t

Sfinal = 50t + 1/2 (a * t * t)
Sfinal = 50t + 1/2 (-50t)
Sfinal = -25t

Going back to setting each equation equal to each other
-25t = 20t + 100
-45t = 100
t = -20/9

To the original USS equation:
Sfinal = Sinitial + Vinitial * t + 1/2 * a * t^2
-25(-20/9) = 0 + 50 * (-20/9) + 1/2 * a * (-20/9)^2
500/9 = -1000/9 + 1/2 * a* 400/81
a = -67.5 km/s^2
a = -6750 m/s^2

Which is the wrong answer.

The correct answer is -4500 m/s^2. I'm only 150% off

Help please?

 

[Random Variable]

It seems like a problem involving general relativity since you have an accelerating reference frame and relatively high speeds.

 

[MrCodeDude]

Originally posted by: Random Variable
It seems like a problem involving general relativity since you have an accelerating reference frame and relatively high speeds.

?

 

[Random Variable]

Originally posted by: MrCodeDude

Originally posted by: Random Variable
It seems like a problem involving general relativity since you have an accelerating reference frame and relatively high speeds.

?

I probably don't know what I'm talking about. I'm only familiar with special relativity.

EDIT: And 50km/s is less than 30% of the speed of light. Just ignore me.

 

[dighn]

basically you want relative distance to be > 0 when relative speed becomes 0

relative speed = 30
relative distance = 100

time for speed to become 0 = 30/a
distance = -1/2at^2 + vit + di = -450/a + 900/a + 100 = 0

a = 4.5 km/s^2

actally lemmie fix the signs:

30+at = 0
t = -30/a

distance = 1/2at^2 + vit + di = 450/a -900/a - 100 > 0
-450/a > 100
a < -4.5 km/s^2 = 4500 m/s^2

i can't believe i messed up the signs against... too tired

 

[ucdbiendog]

try using the final velocity of 20 km/s for the USS enterprise. that will "barely" avoid collision.

 

[PurdueRy]

No problem. Here you go:

You can treat the slower ship as if it is standing still, the other ship would be approaching with a speed of 50 - 20 = 30Km/s

so

Vo^2=Vf^2+2ax
(30,000m/s)^2 = 0^2+2a(100,000m)
a = 4500 and since it is acting against the movement, -4500 m/s^2 is the answer

 

[MrCodeDude]

Originally posted by: PurdueRy
No problem. Here you go:

You can treat the slower ship as if it is standing still, the other ship would be approaching with a speed of 50 - 20 = 30Km/s

so

Vo^2=Vf^2+2ax
(30,000m/s)^2 = 0^2+2a(100,000m)
a = 4500 and since it is acting against the movement, -4500 m/s^2 is the answer

Jesus christ that's easy. I was trying to make it really hard

Wow, you guys from Purdue are great. I asked on ATOT cause agnitrate was sleeping

:heart: Purdue (except when they play tOSU)

 

[MrCodeDude]

Originally posted by: dighn
basically you want relative distance to be > 0 when relative speed becomes 0

relative speed = 30
relative distance = 100

time for speed to become 0 = 30/a
distance = -1/2at^2 + vit + di = -450/a + 900/a + 100 = 0

a = 4.5 km/s^2

actally lemmie fix the signs:

30+at = 0
t = -30/a

distance = 1/2at^2 + vit + di = 450/a -900/a + 100 > 0
-450/a > -100
a < -4.5 km/s^2 = 4500 m/s^2

:heart: Much love for you too

 

[PurdueRy]

Originally posted by: MrCodeDude

Originally posted by: PurdueRy
No problem. Here you go:

You can treat the slower ship as if it is standing still, the other ship would be approaching with a speed of 50 - 20 = 30Km/s

so

Vo^2=Vf^2+2ax
(30,000m/s)^2 = 0^2+2a(100,000m)
a = 4500 and since it is acting against the movement, -4500 m/s^2 is the answer

Jesus christ that's easy. I was trying to make it really hard

Wow, you guys from Purdue are great. I asked on ATOT cause agnitrate was sleeping

:heart: Purdue (except when they play tOSU)

Then no worries this year ...not that you would have to be with the way our team is playing. And yeah, its pretty easy once you recognize what you can do with the velocities.