The Starship Enterprise returns from warp drive to ordinary space with a forward

Question

The Starship Enterprise returns from warp drive to ordinary space with a forward speed of 50 km/s. To the crew's great surprise, a Klingon ship is 150 km directly ahead, traveling in the same direction at a mere 22 km/s. Without evasive action, the Enterprise will overtake and collide with the Klingons in just about 5.4 s. The Enterprise's computers react instantly to brake the ship.


Part A
What magnitude acceleration does the Enterprise need to just barely avoid a collision with the Klingon ship? Assume the acceleration is constant.



I understand how to do most of the problem, I am just getting stuck for time.

Explanation / Answer

Hi, Let 'a' be the acceleration applied by the enterprise to slow down. Since the Klingons is moving with a consant spped of of 22km/s = 22 * 10^3 m/s, we need to slow down the speed of our enterprise also to that velocity just the moment it reaches the klingons in a time say 't'. In this same time t, klingsons will travel a distance of say 'x' and so the enterprise will travel a distance of 'x+150' km. Now we can solve this by considering the equations of displacement for both of them and the final velocity eq of enterprise. However, these kind of problems can also be solved much easier using the relative velocity concept. Initial velocity of enterprise = vie = 50km/s = 50000 m/s. Initial velocity of the klingsons = vik = 22km/s = 22000km/s. Both moving in the same direciton. Hence the resultant velocity = vi = vie - vik                                                                   = 50000 - 22000 = 28000 m/s. And the difference in the final velocities will be vf = vfe - vfk = 22000 -22000 = 0m/s       (fromt the argument above, its enough that both should be travelling at same speed). Hence the situation can be reconsidered as enterprise moving with initial velocity of 28000 m/s towards klingsons which is at rest and at a distance of 150km = 150000m. So now we need the acceleration that brings the enterprise moving with the above initial velocity to rest at 150000 m. From equation vf^2  - vi^2 = 2 * a * s.                         a = - vi^2 / 2*s                            = - (28000)^2 / (2 * 150000)                               = - 2613.33 m/s^2 (-ve sign implies that its decceleration) Hope this helps you. Hope this helps you.

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