A spacecraft is moving in space, where all the external forces can be neglected. Any change in its s

Q: Why can't the spacecraft just turn instantly?

An instant turn would require an infinite change in velocity in zero time, which implies infinite acceleration. The spacecraft's rockets have a maximum acceleration limit of $5\sqrt{2} \text{ m/s}^2$, so the turn must take some finite time.

Q: Why must the acceleration be constant in direction?

The change in velocity is the vector sum (integral) of acceleration over time. To get the largest possible change in velocity in a given time, all the tiny acceleration vectors must point in the exact same direction. If the acceleration changed direction, some of its effort would be wasted 'canceling out' previous changes, taking more time.

Q: Why is the path a parabola and not a circle?

A circular path requires the acceleration (centripetal) to constantly change direction so it always points towards the center. In this problem, to minimize time, we found that the acceleration must remain constant in both magnitude AND direction. Constant acceleration with an initial velocity at an angle always results in a parabolic trajectory, just like a thrown ball under gravity.

Q: How did we get $\Delta \vec{v} = v\sqrt{2}$?

The initial velocity $\vec{v}_i$ and final velocity $\vec{v}_f$ are perpendicular (a $90^\circ$ turn) and have the same magnitude $v$. The change in velocity is $\Delta \vec{v} = \vec{v}_f - \vec{v}_i$. Geometrically, this forms a right-angled triangle where $\Delta \vec{v}$ is the hypotenuse. By Pythagoras, $|\Delta \vec{v}| = \sqrt{v^2 + v^2} = \sqrt{2v^2} = v\sqrt{2}$.

Q: Does the spacecraft actually stop during the turn?

No, the spacecraft never stops. Its speed changes during the turn, but it never drops to zero. The velocity vector smoothly transitions from the initial direction to the final direction while tracing the parabolic path.

Question

A spacecraft is moving in space, where all the external forces can be neglected. Any change in its speed and direction of motion can be accomplished by rockets installed on it. At an instant when it is moving with a speed $v = 100 	ext{ m/s}$, the crew inside decides to take a $90^\circ$ turn with an acceleration of constant modulus and then move in the new direction with the same speed $v$. The rockets installed can provide a maximum acceleration $a = 5\sqrt{2} 	ext{ m/s}^2$. Find the minimum time spent and shape of the path followed during the turn.

Sigma Percentile · Accepted Answer

### The Spacecraft's Dilemma

Imagine you are the pilot of a spacecraft cruising through the frictionless void of deep space at a constant speed of $v = 100 	ext{ m/s}$. Suddenly, mission control orders an immediate $90^\circ$ turn. The catch? You must exit the turn at the exact same speed of $100 	ext{ m/s}$, and you must complete this maneuver in the absolute minimum time possible. Your thrusters can provide a maximum acceleration of $a = 5\sqrt{2} 	ext{ m/s}^2$.

How do you orient your thrusters, how long will it take, and what path will your ship carve through the stars?

### Decoding the Velocity Vectors

To solve this, we must first understand exactly what is changing. Velocity is a vector, meaning it has both magnitude (speed) and direction.

Let's set up a coordinate system. Assume your initial velocity $\vec{v}_i$ is directed along the positive x-axis. Therefore, $\vec{v}_i = 100 \hat{i}$. After the $90^\circ$ turn, your final velocity $\vec{v}_f$ will be along the positive y-axis, so $\vec{v}_f = 100 \hat{j}$.

The total change in velocity required for this maneuver is denoted by $\Delta \vec{v}$. By definition:

$$\Delta \vec{v} = \vec{v}_f - \vec{v}_i = 100 \hat{j} - 100 \hat{i}$$

To find the magnitude of this required change, we use the Pythagorean theorem:

$$|\Delta \vec{v}| = \sqrt{(-100)^2 + (100)^2} = 100\sqrt{2} 	ext{ m/s}$$

This $100\sqrt{2} 	ext{ m/s}$ is the total 'velocity distance' your thrusters need to cover.

### The Physics of Minimum Time

Now comes the crucial optimization step. We know from kinematics that the change in velocity is the integral of acceleration over time:

$$\Delta \vec{v} = \int \vec{a} \, dt$$

To achieve a specific $\Delta \vec{v}$ in the **minimum possible time**, two conditions must be met:
1. The acceleration must be at its maximum possible magnitude at all times ($|\vec{a}| = a_{max}$). 
2. The acceleration vector must **constantly point in the exact direction** of the required $\Delta \vec{v}$.

If the acceleration vector were to change direction, some of its effort would be wasted pushing the ship in directions that don't contribute directly to the final goal, thereby increasing the time taken.

### Calculating the Maneuver Time

Since we have established that the optimal acceleration $\vec{a}$ is constant in both magnitude and direction, the integral simplifies beautifully to a basic algebraic equation:

$$|\Delta \vec{v}| = a_{max} \cdot t_{min}$$

We can now solve for the minimum time $t_{min}$:

$$t_{min} = \frac{|\Delta \vec{v}|}{a_{max}}$$

Substituting our known values:

$$t_{min} = \frac{100\sqrt{2}}{5\sqrt{2}} = 20 	ext{ s}$$

The maneuver will take exactly 20 seconds.

### Tracing the Cosmic Path

Finally, what is the geometric shape of the path the spacecraft follows during these 20 seconds?

Let's review the conditions we've established:
- The spacecraft has an initial velocity $\vec{v}_i$.
- It is subjected to a **constant acceleration vector** $\vec{a}$.
- The acceleration vector $\vec{a}$ (which points in the direction of $\Delta \vec{v}$) is **not parallel** to the initial velocity $\vec{v}_i$.

In classical mechanics, whenever a particle moves under the influence of a constant acceleration that is at an angle to its initial velocity, the resulting trajectory is always a **parabola**.

Think of a ball thrown horizontally on Earth. Gravity provides a constant downward acceleration, while the ball has an initial horizontal velocity. The result is a familiar parabolic arc. Our spacecraft experiences the exact same kinematic conditions, just in the vacuum of space. Therefore, it will trace a smooth parabolic path as it executes its optimal 20-second turn.

Visualized Solution

Initial and Final Velocities

Change in Velocity $Δ v$

Magnitude of $Δ v$

Condition for Minimum Time

Calculating Minimum Time

Substituting Values

Shape of the Path

Final Conclusion

The Sigma Insight: Motion in a Plane

The Spacecraft's Dilemma

Decoding the Velocity Vectors

The Physics of Minimum Time

Calculating the Maneuver Time

Tracing the Cosmic Path

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