Question

# The Starship Enterprise returns from warp drive to ordinary space with a forward speed of 56...

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.

Solution:

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 km/s2

Check :

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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