Question

Review 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 120 km directly ahead, traveling in the same direction at a mere 27 km/s. Without evasive action, the Enterprise will overtake and collide with the Klingons in just about 4.3 S. The Enterprises 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 Hint: Draw a be the location of the Ent -versus-time graph showing the motions of both the Enterprise and the Klingon ship. Let ro okm 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 graphicaly Express your answer to two significant figures and inelude the appropriate units. aValue Units

Answer 1

Forward speed = 55 k m/s Speed of Klingon Ship = 27 km/s at distance 120 km ahead time = 4.3 Let we denote x = starship y = Klingon Now average velocity of starship, V_x = V_x_i + V_x_f/2 Where V_x_f = 0 (because it stop) Therefore V_x = (55 + 0)/2 = > 27.5 km/s Distance travel by x before stop = 120 km + distance of y in the given time. Now (d)_x = (d)_y + 120 (27.5)t = 27t + 120 0.5t = 220 t = 240 s S, (d)_x = (27.5) (240) = > 6600 km Now (d)_x = u_x_i t + 1/2 at^2 6600 = (55) (240) + 1/2 a (240)^2 6600 = 13200 + 1/2 times a times 57600 a = -6600/28800 = > -0.229 km/s^2 a = -2.3 times 10^-1 km/s^2