KINEMATICS BYU 38

Solution to Question 38

Problem Analysis:
The spacecraft needs to change its velocity vector from $\vec{v}_i$ to $\vec{v}_f$ such that $|\vec{v}_i| = |\vec{v}_f| = v$ and the angle between them is $90^\circ$. The acceleration magnitude $a$ is constant.

1. Calculating Change in Velocity ($\Delta \vec{v}$):
Let $\vec{v}_i = v \hat{i}$ and $\vec{v}_f = v \hat{j}$.
$$ \Delta \vec{v} = \vec{v}_f – \vec{v}_i = v \hat{j} – v \hat{i} $$ The magnitude of the required change is: $$ |\Delta \vec{v}| = \sqrt{v^2 + (-v)^2} = v\sqrt{2} $$

2. Minimizing Time:
To achieve this change in the minimum time $t$, the acceleration vector $\vec{a}$ must be directed constantly along the direction of the velocity change vector $\Delta \vec{v}$. If the acceleration direction rotates, it would be inefficient.

Using the kinematic equation $\Delta \vec{v} = \vec{a} t$ (scalar form for constant direction): $$ t = \frac{|\Delta \vec{v}|}{a} = \frac{v\sqrt{2}}{a} $$ Given $v = 100$ m/s and $a = 5\sqrt{2}$ m/s²: $$ t = \frac{100\sqrt{2}}{5\sqrt{2}} = 20 \text{ s} $$

3. Shape of the Path:
Since the acceleration vector $\vec{a}$ is constant in both magnitude and direction throughout the turn, the motion is equivalent to projectile motion (under a constant force field like gravity). The path of an object moving with an initial velocity and subjected to a constant force (acceleration) at an angle is a Parabola.