- Homework Statement:
The Starship Enterprise returns from warp drive to ordinary
space with a forward speed of 50 km/s. To the crew’s great sur- prise, a Klingon ship is 100 km directly ahead, traveling in the same direction at a mere 20 km/s. Without evasive action, the Enterprise will overtake and collide with the Klingons in just slightly over 3.0 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.
- Relevant Equations:
vf^2 = vi^2 + 2at
xf = xi + vt
I drew a x(t) graph to try to map out what was going on. I guessed that in order to just barely avoid the crash, the velocity of Enterprise (v.e) will have to decrease to match the velocity of Klingon (v.k). So v.e final = v.k
Since we're looking for the acceleration, I used this formula:
vf^2 = vi^2 + 2a∆x
v.k^2 = v.e^2 + 2a∆x
for ∆x, I looked for the position of Klingon at t = 3 s.
xf = xi + vt
xf = 100 + 20(3)
xf = 160
∆x would then => 160 - 0 = 160 so
20^2 = 50^2 + 2a(160)
a = -6.6 km/s
which is obviously wrong :( the answer is supposed to be 4500 m/s^2
I feel like I'm missing a fundamental understanding of this problem. Will the Enterprise's speed not be equal to the speed of Klingon at t=3s? I'm super confused.
*I know someone asked about this question already, but I unfortunately still did not understand :( Thank u for your patience