Question

The Starship Enterprise returns from warp drive to ordinary space with a forward speed of 60 km/s . To the crew's great surprise, a Klingon ship is 140 km directly ahead, traveling in the same direction at a mere 30 km/s . Without evasive action, the Enterprise will overtake and collide with the Klingons in just about 4.7 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.
Hint: Draw a position-versus-time graph showing the motions of both the Enterprise and the Klingon ship. Let x0 = 0km be the location of the Enterprise 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 graphically, express that situation mathematically.

Express your answer to two significant figures and include the appropriate units.

a=(value)(units)

Answer 1

[please note: if the the velocities of objects A(Enterprise) and B(Klingon) are v? and v? with respect to a stationary observer (neutral observer) on the ground surface, then the velocity of A relative to B is v? - v?. this is as if B were at rest and only A were moving.]

initial v? - v? = 60 - 30 = 30 km/s

relative acceleration = a? - 0 = a? km/s²

this is as if ship B is at rest(140 km away) and ship A is moving with an initial velocity 30 km/s and acceleration 'a?' km/s² towards it.

relatively applying v² = u² + 2as

=> 0 = 30² + 2a?(140)

or a? = -3.214 km/s² (a deceleration)

So magnitude of acceleration is:

a? = 3.2 km/s²