What is the maximum speed achieved by a spaceship traveling to Moon Base I?

In summary, the spaceship travels a straight line path from Earth to Moon Base I with an acceleration of 19.4 m/s2 for the first 14.4 minutes and then travels at a constant speed until the last 14.4 minutes when it decelerates to rest at the moon. The maximum speed attained can be found using the formula for the area of a trapezoid. The fraction of the total distance traveled at constant speed can be calculated using a v-t graph. The total time required for the trip can be found by converting the time in minutes to seconds. Graphing can be a helpful tool in solving this type of problem.
  • #1
azila
67
0
Another hard physics problem??

Homework Statement


A spaceship ferrying workers to Moon Base I takes a straight line path from the Earth to the moon, a distance of 384000 km. Suppose it accelerates at an acceleration 19.4 m/s2 for the first time interval 14.4 min of the trip, then travels at constant speed until the last time interval 14.4 min when it accelerates at 19.4 m/s2 just coming to rest as it reaches the moon.
A. What is the maximum speed attained

B. What fraction of the total distance is traveled at constant speed

C. What total time is required for the trip

Homework Equations


So, I made a table
X Vo V A t
First Time interval 2011.39 0 279.36 19.4 14.4
Constant Speed 379977 279.36 ? ? ?
Last time interval 2011.39 ? ? 19.4 14.4

The Attempt at a Solution



First of all, is acceleration 0 when velocity is constant? If it is then in the constant speed time, the acceleration is zero and then the velocity would be the same. But I don't know how to get the max speed attained. What should I do? I tried using every one of the constant acceleration equations. When I tried solving for time, it was giving me 22 hours and that is not the answer. So please, if you can help me a bit out. Thanks
 
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  • #2
azila said:

Homework Statement


A spaceship ferrying workers to Moon Base I takes a straight line path from the Earth to the moon, a distance of 384000 km. Suppose it accelerates at an acceleration 19.4 m/s2 for the first time interval 14.4 min of the trip, then travels at constant speed until the last time interval 14.4 min when it accelerates at 19.4 m/s2 just coming to rest as it reaches the moon.
A. What is the maximum speed attained

B. What fraction of the total distance is traveled at constant speed

C. What total time is required for the trip

Homework Equations


So, I made a table
X Vo V A t
First Time interval 2011.39 0 279.36 19.4 14.4
Constant Speed 379977 279.36 ? ? ?
Last time interval 2011.39 ? ? 19.4 14.4

The Attempt at a Solution



First of all, is acceleration 0 when velocity is constant? If it is then in the constant speed time, the acceleration is zero and then the velocity would be the same. But I don't know how to get the max speed attained. What should I do? I tried using every one of the constant acceleration equations. When I tried solving for time, it was giving me 22 hours and that is not the answer. So please, if you can help me a bit out. Thanks

yes, the acceleration is zero when the velocity is constant.

You seemed to have made a mistake with the units. You must put the time in seconds if you use an acceleration in m/s^2
 
  • #3
it's good to make a graph for these types of questions..
then you really don't need to know any of those formulas ^^

for this, a v-t graph can help a lot

"I don't know how to get the max speed attained. What should I do? I tried using every one of the constant acceleration equations. When I tried solving for time, it was giving me 22 hours and that is not the answer. So please, if you can help me a bit out. Thanks"

ignore everything else, but this statement: " Suppose it accelerates at an acceleration 19.4 m/s2 for the first time interval 14.4 min of the trip, "
answer to your first question

Hint(if you trying graph way): Area of trapezoid = 0.5*v*(t when v is contant)*(t when v is contant+2*14.4)=total distance
 
Last edited:
  • #4
Ok, thanks for telling me to convert the time in seconds because the time needs to be in seconds for the equation to work. Thanks rootx for the formula, I used that to get it. Yeah, graphing really helps. Thanks again for every ounce of help.
 

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