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.

To solve this problem, let us say that

A = starship enterprise

B = klingon ship

Assuming that the starship enterprise comes to a constant deceleration, then the average velocity would be:

vA = (vAi + vAf) / 2                 where vAf = 0 since it must come to a stop

vA = (50 + 0)/2

vA = 25 km/s

Therefore the distance that A must travel before it comes to a stop is 100 km plus the distance of B that it travels within that time: 
dA = dB + 100

25t = 20t + 100 

5t = 100 
t = 20 s

 
So 
dA = 25t = 25(20) = 500 km 

and 

dA = vAi * t + 1/2 a t^2 
500 = 50(20) + 1/2 a (20)^2 
a = -2.5 km/s^2          

So the enterprise needs to decelerate at -2.5 km/s^2 to avoid collision. 

+15-35=-20

going in opposite directions

answer( sicar );