Starship Enterprise and Klingon Ship Problem on Physics| University Physics

The Starship Enterprise returns from warp drive to ordinary space with a forward speed of 51 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 5.0 s. The Enterprise’s computers react instantly to brake the ship.

PART A. What magnitude acceleration does the Enterprise need to just barely avoid a collision with the Klingon ship? Assume the acceleration is constant.

ANSWER: a=3.0\:km/s^2

Since the collision is barely avoided, we shall make their positions and their velocities equal. So, their velocities must be equal to the Klingon ship. For the Starship Enterprise, the final velocity will be equal to 21 km/s

a=\frac{v_f-v_i}{\Delta t}

a=\frac{21-51}{t}

t=-\frac{30}{a}

Equate their positions as well

x_{enterprise}=x_{ship}

51t+\frac{1}{2}at^2=150+21t

Substitute t=-\frac{30}{a} to all t’s

150-30\left(-\frac{30}{a}\right)=\frac{1}{2}a\left(-\frac{30}{a}\right)^2

150+\frac{900}{a}=\frac{450}{a}

\frac{450}{a}=-150

a=-\frac{450}{150}

a=-3.0\:km/s^3

Therefore, the magnitude of the acceleration should be a=3.0\:km/s^3

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