Star ship acceleration physics

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 6K views
ms245004
Messages
4
Reaction score
0

Homework Statement


The Starship Enterprise returns from warp drive to ordinary space with a forward speed of 56km/s. To the crew's great surprise, a Klingon ship is 120m directly ahead, traveling in the same direction at a mere 24km/s. Without evasive action, the Enterprise will overtake and collide with the Klingons in just about 3.8s. The Enterprise's computers react instantly to brake the ship. What magnitude acceleration does the Enterprise need to just barely avoid a collision with the Klingon ship? Assume the acceleration is constant.

Homework Equations


Since the acceleration is constant I know you use one of the three kinematic formulas but other then that I am lost. Please help me.


The Attempt at a Solution

 
Physics news on Phys.org
You have the initial conditions. The trick to solving this problem is interpreting the phrase, "just barely avoid a collision with the Klingon ship" so that you know what the final conditions are supposed to be. In this case, the phrase should be interpreted to mean that at the instant the Enterprise "catches up" to the Klingon vessel, its velocity has been reduced to zero, so that there is no real impact. In other words, the Enterprise slows to a stop just in time (at the instant at which the two ships barely make contact).

"Catches up" means that their positions are the same, right? So you'll have two equations: one for the position vs. time of the Klingon ship, which travels at constant speed, and one for the position vs. time of the Enterprise, which travels at constant acceleration.

List the quantities you have been given in the problem, as well as the quantities you have to determine, (the "unknowns"). That will help you to figure out what physics is required to do so.
 
So for Klingon I used sf = si + v(tf - ti) and got the sf equals 211.2 Then using that and the equation sf = si + vi(tf-ti) + .5a(tf-ti)^2 and solved for a getting -.221. When I put that in as an answer i am told it is wrong. What am i still doing wrong here?
 
ms245004 said:
So for Klingon I used sf = si + v(tf - ti) and got the sf equals 211.2 Then using that and the equation sf = si + vi(tf-ti) + .5a(tf-ti)^2 and solved for a getting -.221. When I put that in as an answer i am told it is wrong. What am i still doing wrong here?

I am assuming that you used 3.8 s as the time interval tf-ti. Can you see why this is obviously wrong? The 3.8 s is irrelevant information that just happens to have been included in the problem. It is the time it would take for them to collide if the Enterprise did not take any action to avoid the collision. However, the fact is, the Enterprise DOES take action to avoid the collision by slowing down with a constant acceleration. Therefore, you don't know what tf is from beforehand -- you have to figure it out by applying the conditions that must be true, namely:

- At tf, the positions of the two ships are the same
- At tf, the velocity of the Enterprise is 0.
 
Last edited: