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 110 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 about 2.8 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. Express your answer to two significant figures and include the appropriate units. a = _______

Answer 1

The deceleration that Enterprise needs to make should be such that when it reaches Kingon's position, its speed matches Kingon's speed. Also, the duration of deceleration should be such that the above is achieved.

Let the duration of acceleration be 't' and the deceleration be 'a'

Intial velocity of Enterprise = 60 km/s=60000m/s

Final velocity of Enterprise = 20km/s=20000m/s

20000=60000+at

Or, a=-40000/t....(i)

Also, distance travelled by Enterprise in time t = 60000t+(1/2)at²

Distance travelled by Kingon in the same duration=20000t

60000t+(1/2)at² - 20000t=110000

Or, 40000t+(1/2)at² - 110000=0 .....(ii)

Substituting a=-40000/t from (i) in (ii), we get

40000t+(1/2)(-40000/t)t² - 110000=0

Or , 20000t=110000

t = 110000/20000=5.5 seconds

a =  a=-40000/t = -40000/5.5=-7272.73 m/s² = -7.27 km/s²

Therefore, Enterprise needs to decelerate at 7.27 km/s²

The position(y axis) vs time(x axis) graphs of both Enterprise(blue) and Kingon (Red) are as shown below. To barely avoid ollision, the graphs of Enterprise and Kingon should be such that they barely touch each other. If the graphs touch each other at some time and position, this means both are at same location and collision tales place.

The graph for Enterprise has been drawn with acceleration =-7272.72 m/s