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 55 km/s . To the crew's great surprise, a Klingon ship is 150 km directly ahead, traveling in the same direction at a mere 30 km/s . Without evasive action, the Enterprise will overtake and collide with the Klingons in just about 6.0 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. Hint. Draw a position-versus-time graph showing the motions of both the Enterprise and the Klingon ship. Let x_0 = 0km be the location of the Enterprise as it returns from warp drive. How do you show graphically the situation in which the collision is "barely avoided"? Once you decide what it looks like graphically, express that situation mathematically. Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

The thing to keep in mind here is that the collision doesn't occur at the point 150 km away. Instead,
Enterprise: s = vo*t + ½at² = 55km/s*t + ½at²
Klingon: S = So + Vo*t = 150km + 30km/s*t

Enterprise must drop its speed to 30 km/s, so
30km/s = 55km/s + at
a = -25km/s / t plug that into displacement equation:
s = 55km/s*t + ½at² = 55km/s*t + ½(-25km/s / t)t²
s = 55km/s*t - 12.5km/s*t = 42.5km/s * t
and now set s = S
42.5km/s * t = 155km + 30km/s*t
t = 155km / 12.5km/s = 12.4 s
(Note that the statement "Without evasive action, the Enterprise will overtake and collide with the Klingons in just about 6.2 s" is patently incorrect. It would be true if the Klingons were stationary, but they aren't.)
Then a = v / t = -25km/s /12.4s = -2.016 km/s² acceleration

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