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.