# Starship Enterprise and Klingon Ship Problem on Physics| University Physics

### 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 $a=3.0\:km/s^3$