Calculating Maximum Speed Attained by a Spaceship Traveling to the Moon

[Whatupdoc]

A spaceship ferrying workers to Moon Base I takes a straight-line path from the Earth to the moon, a distance of 384,000 km. Suppose it accelerates at an acceleration 19.8 $$m/s^2$$ for the first time interval 15.9 min of the trip, then travels at constant speed until the last time interval 15.9 min, when it accelerates at -19.8 $$m/s^2$$, just coming to rest as it reaches the moon.

there are three questions to this problem, but i will ask the first one first.

1.)What is the maximum speed attained?

using the formula d1 = x(0) +v(0)*t + 1/2at^2

x(0) = 0 because the velocity is zero
d1 = 1/2(19.8) * (15.9 *60)^2 <--- converted to secs.

well that's the speed for distance #1, but since it's constant just before it reaches distance #3, shouldn't that be the maximum speed?

 

[dink]

I believe you only need to use V(final) = V(intial) + at.

Since intial is 0? then your maximum speed would be acceleration * time interval of 15.9min? So converting that to seconds gives (19.8 m/s^2) * (15.9 min * 60 s/m).

 

[needhelpperson]

is this even a valid college level question?

 

[dink]

Yes, that is first year college mechanics.

 

[Whatupdoc]

I believe you only need to use V(final) = V(intial) + at.

Since intial is 0? then your maximum speed would be acceleration * time interval of 15.9min? So converting that to seconds gives (19.8 m/s^2) * (15.9 min * 60 s/m).

(19.8 m/s^2) * (15.9 min * 60 s/m) = 18889.2 and it's the wrong answer


and i don't know what you mean by if it's a college question, but i am in college. it may seem easy because it's only been the first week of school.

 

[dink]

Your asking the maximum speed attained, which is the magnitude of the velocity. From what I gather of the problem, the ship has positive acceleration for a period of 15.9 minutes, a constant velocity, then a negative acceleration for a period of 15.9 minutes. Your equation is the distance equation which, if you look at the units, leaves you an answer in meters. Looking for a maximum velocity will have the units m/s. Regardless of the results the equation is most assuredly Vi = Vf + AT.

 

[Tide]

There's a serious flaw in the problem. If the rocket is traveling a straight line path to the moon then part of that acceleration is required to keep it on the straight line path to compensate for the varying angular momentum on its way to the moon. I think the problem needs to be restated!

 

[Whatupdoc]

Your asking the maximum speed attained, which is the magnitude of the velocity. From what I gather of the problem, the ship has positive acceleration for a period of 15.9 minutes, a constant velocity, then a negative acceleration for a period of 15.9 minutes. Your equation is the distance equation which, if you look at the units, leaves you an answer in meters. Looking for a maximum velocity will have the units m/s. Regardless of the results the equation is most assuredly Vi = Vf + AT.

answer is suppose to be in km/s. sorry, forgot to state that. so am i suppose to times it by 1000, cause the original answer of 18889.2 is in meters right?

and Tide, it's a copy and paste from my homework(i didnt type it).

 

[Tide]

and Tide, it's a copy and paste from my homework(i didnt type it).

I was only suggesting that whoever made up the problem was somewhat sloppy!

 

[Chronos]

The formula Dink gave is correct and 18889 is the right answer. Given, the deceleration phase is of the same magnitude and duration as the launch, there can be no other answer [unless the ship crashes].

 

[Whatupdoc]

is 18889 in meters? the answer is suppose to be in km/s. so do i convert 18889 meters to ____km/s?

 

[needhelpperson]

...sigh...

 

[HallsofIvy]

The formula Dink gave is correct and 18889 is the right answer. Given, the deceleration phase is of the same magnitude and duration as the launch, there can be no other answer [unless the ship crashes].

The formula is correct but not for this question!

1.)What is the maximum speed attained?

using the formula d1 = x(0) +v(0)*t + 1/2at^2

This is the formula for distance, not speed!

The formula for speed is simply v(0)t.

19.8*15.9= 314.82 m/s for the maximum speed.

 

[Whatupdoc]

The formula is correct but not for this question!



This is the formula for distance, not speed!

The formula for speed is simply v(0)t.

19.8*15.9= 314.82 m/s for the maximum speed.

314.82 m/s is also the wrong answer

 

[needhelpperson]

A spaceship ferrying workers to Moon Base I takes a straight-line path from the Earth to the moon, a distance of 384,000 km. Suppose it accelerates at an acceleration 19.8 $$m/s^2$$ for the first time interval 15.9 min of the trip, then travels at constant speed until the last time interval 15.9 min, when it accelerates at -19.8 $$m/s^2$$, just coming to rest as it reaches the moon.

there are three questions to this problem, but i will ask the first one first.

1.)What is the maximum speed attained?

using the formula d1 = x(0) +v(0)*t + 1/2at^2

x(0) = 0 because the velocity is zero
d1 = 1/2(19.8) * (15.9 *60)^2 <--- converted to secs.

well that's the speed for distance #1, but since it's constant just before it reaches distance #3, shouldn't that be the maximum speed?

vf = 19.8*15.9*60 = 18889.2m/s -> DIVIDE BY 1000! TO GET KM/S. And don't tell me that its your first week of college, which is why even this simple conversion is too hard for you.

answer is suppose to be in km/s. sorry, forgot to state that. so am i suppose to times it by 1000, cause the original answer of 18889.2 is in meters right?

and Tide, it's a copy and paste from my homework(i didnt type it).

18.889.2 km/s

 

[Whatupdoc]

i never took physics in high school, I am just trying to learn. all of this is so new and hard. in college, a lot of stuff is given to you at once and they all go so fast. classes are huge, so it's hard to ask questions(around 250 or more students in my physics class).

 

[Chronos]

The 18889 was in meters, as I'm sure you now realize. Hehe, shame on you Halls, you forgot to multiply the minutes by 60.