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 58km/s . To the crew's great surprise, a Klingon ship is 120km directly ahead, traveling in the same direction at a mere 30km/s . 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.

here is my work, what am i doing wrong

58/2 = 29

29t=30t+120

-1t=120

t=120

29(-120)=-3480km

-3480=58(-120) + (1/2)a(-120)^2

3480=1/2a(-120)^2

a=?????

Explanation / Answer

so to just barely miss a collision when want x klingon = x enterprise when v klingon = venterprise so v equation tells us v = v0 + a t 30 = 58 + at t = -28/a x equation x klingon = x enterprise 120 + 30 t = 0 + 58 t + 1/2 a t^2 120 + 30*-28/a = 58*-28/a + 1/2 a*(-28/a)^2 solve for a a= -3.27 km/s^2

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