To solve this problem, we need to determine the magnitude of the acceleration required for the Enterprise to just barely avoid colliding with the Klingon ship. We can use kinematic equations to model the motion of both ships and find the necessary acceleration.

Given Data:

• Initial speed of the Enterprise, $v_{\text{Enterprise}} = 60 \, \text{km/s}$
• Speed of the Klingon ship, $v_{\text{Klingon}} = 25 \, \text{km/s}$
• Distance between the two ships initially, $d = 150 \, \text{km}$
• Time until collision without braking, $t = 4.3 \, \text{s}$

Step 1: Set up the kinematic equations

For both the Enterprise and the Klingon ship, we will use the equation for position as a function of time, given by: $x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$ Where:

• $x(t)$ is the position at time $t$,
• $x_0$ is the initial position,
• $v_0$ is the initial velocity,
• $a$ is the constant acceleration.

For simplicity, we'll assume that both ships start at $t = 0$, and we want to find the acceleration needed for the Enterprise to stop before it reaches the Klingon ship at $t = 4.3$ s.

Position of the Klingon ship:

The Klingon ship travels with constant speed ($a = 0$), so its position as a function of time is: $x_{\text{Klingon}}(t) = 150 + 25t$ where 150 km is the initial distance between the ships.

Position of the Enterprise:

The Enterprise starts at $x_0 = 0$ km, with an initial speed of 60 km/s. It will decelerate with an unknown acceleration $a$, so its position is: $x_{\text{Enterprise}}(t) = 60t + \frac{1}{2} a t^2$ We need the Enterprise to decelerate such that it reaches the position of the Klingon ship (150 km ahead) at exactly $t = 4.3$ s.

Step 2: Set the positions equal to avoid collision

To avoid a collision, the positions of both ships must be equal at $t = 4.3$ s. Therefore, we set the two position equations equal to each other: $60t + \frac{1}{2} a t^2 = 150 + 25t$ Substitute $t = 4.3$ s into the equation and solve for $a$.

$60(4.3) + \frac{1}{2} a (4.3)^2 = 150 + 25(4.3)$

First, calculate the known terms:

$60(4.3) = 258$ $25(4.3) = 107.5$ $(4.3)^2 = 18.49$ So the equation becomes: $258 + \frac{1}{2} a (18.49) = 150 + 107.5$ Simplify: $258 + 9.245a = 257.5$ Subtract 258 from both sides: $9.245a = -0.5$ Solve for $a$: $a = \frac{-0.5}{9.245} = -0.054 \, \text{km/s}^2$

Step 3: Conclusion

The magnitude of the acceleration needed for the Enterprise to avoid the collision is: $\boxed{0.054 \, \text{km/s}^2} \quad \text{or} \quad 54 \, \text{m/s}^2$

This is the required deceleration (negative acceleration) to just barely avoid the collision.

Do you want more details, or have any questions?

5 related questions:

1. How would you solve this problem if the Klingon ship had a changing velocity?
2. What happens if the Enterprise accelerates instead of decelerating?
3. How would the solution change if the initial distance were different?
4. What would the graph of position vs. time look like for both ships?
5. How could you calculate the time to collision without assuming constant acceleration?

Tip: When solving kinematic problems, make sure to use consistent units and carefully check signs for acceleration (positive for speeding up, negative for slowing down).