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 56 . To the crew's great surprise, a Klingon ship is 150 directly ahead, traveling in the same direction at a mere 21 . Without evasive action, the Enterprise will overtake and collide with the Klingons in just about 4.3 . 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.

Explanation / Answer

assuming constant acceleration Vav of Enterprise = (Vi+0)/2 = 29 m/s d_enterprise = d_klingon + 110 29 t = 26 t + 110 3 t = 110 t = 36.667 s so d_enterprise = 29 t = 29(36.667) = 1063.3 km and d = Vi t + 1/2 a t^2 1063.3 = 58(36.667) + 1/2 a (36.667)^2 a = -1.582 km/s/s or -1.6 km/s/s

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