The Starship Enterprise returns from warp drive to ordinary space with a forward speed of 56 km/s . To the crew's great surprise, a Klingon ship is 150 km directly ahead, traveling in the same direction at a mere 21 km/s . Without evasive action, the Enterprise will overtake and collide with the Klingons in just about 4.3 s . 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.
Let the velocities of objects A(Enterprise) and B(Klingon) are
va and vb with respect to a stationary
observer (neutral observer) on the ground surface, then the
velocity of A relative to B is va - vb. this
is as if B were at rest and only A were moving.
initial va - vb = 56 - 21 = 35 km/s
relative acceleration = aa - 0 = aa km/s2
this is as if ship B is at rest(150 km away) and ship A is moving with an initial velocity 35 km/s and acceleration 'aa' km/s2 towards it.
relatively applying v2 = u2 + 2as
=> 0 = 352 + 2aa(150)
or aa = -4.08 km/s² (a deceleration)
And Magnitude of aceeration is = 4.1
time taken from v = u + at
t = [0 - 35]/[-4.08] = 8.57 s
in actuality A(Enterprise) covers s = 56(8.57) - 0.5(4.08)(8.57)2 = 330 km
B covers a distance d = 21 *8.57 = 180 km
The difference ~= 150 km
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