Question

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.

Answer #1

**Solution:**

Let the velocities of objects A(Enterprise) and B(Klingon) are
v_{a} and v_{b} with respect to a stationary
observer (neutral observer) on the ground surface, then the
velocity of A relative to B is v_{a} - v_{b}. this
is as if B were at rest and only A were moving.

initial v_{a} - v_{b} = 56 - 21 = 35 km/s

relative acceleration = a_{a} - 0 = a_{a}
km/s^{2}

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 'a_{a}'
km/s^{2} towards it.

relatively applying v^{2} = u^{2} + 2as

=> 0 = 35^{2} + 2a_{a}(150)

or a_{a} = -4.08 km/s² (a deceleration)

And Magnitude of aceeration is = **4.1
km/s ^{2}**

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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