Let's break down and solve the questions provided.

Given:

• Initial velocity $\mathbf{v}_0 = 1000 \, \text{m/s} \, \mathbf{i}$.
• At $t = 0$, the spaceship turns on its thrusters, giving it a constant acceleration $\mathbf{a} = 10 \, \text{m/s}^2 \, \mathbf{j}$ for 10 seconds.
• The acceleration does not change over time.

Questions:

i) Sketch the path of the spaceship without doing any detailed calculations.

Solution:

• Initially, the spaceship has a velocity only in the $\mathbf{i}$ (x-direction), so it is moving horizontally.
• As the thrusters are turned on at $t = 0$, the spaceship experiences a constant acceleration in the $\mathbf{j}$ (y-direction).
• The path of the spaceship will be a parabola due to the combination of the constant horizontal velocity and the constant vertical acceleration.

Sketch: The path is a parabola opening upwards with the x-axis representing time.

ii) Find the velocity of the spaceship as a function of time.

Solution:

The velocity of the spaceship in the x-direction ($\mathbf{i}$) remains constant since there's no acceleration in that direction: $v_x(t) = v_{0x} = 1000 \, \text{m/s} \, \mathbf{i}$

The velocity in the y-direction ($\mathbf{j}$) increases linearly with time due to the constant acceleration: $v_y(t) = a \cdot t = 10 \cdot t \, \mathbf{j}$

So, the total velocity vector $\mathbf{v}(t)$ is: $\mathbf{v}(t) = 1000 \, \mathbf{i} + 10t \, \mathbf{j}$

iii) Find the position $\mathbf{r}(t)$ of the spaceship as a function of time in terms of its initial position $\mathbf{r}(0)$.

Solution:

The position in the x-direction ($\mathbf{i}$) is given by: $x(t) = x(0) + v_{0x} \cdot t$ $x(t) = x(0) + 1000t$

The position in the y-direction ($\mathbf{j}$) is obtained by integrating the velocity in the y-direction: $y(t) = y(0) + \int v_y(t) \, dt$ $y(t) = y(0) + \int 10t \, dt = y(0) + 5t^2$

So, the total position vector $\mathbf{r}(t)$ is: $\mathbf{r}(t) = (x(0) + 1000t) \, \mathbf{i} + \left(y(0) + 5t^2\right) \, \mathbf{j}$

iv) Determine the exact position of the spaceship at time $t = 20$s, assuming an initial position $\mathbf{r}(0) = 50 \, \mathbf{i} + 30 \, \mathbf{j}$.

Solution:

Given $\mathbf{r}(0) = 50 \, \mathbf{i} + 30 \, \mathbf{j}$, substitute into the position function from part (iii):

For the x-component: $x(20) = 50 + 1000 \times 20 = 50 + 20000 = 20050 \, \text{m}$

For the y-component: $y(20) = 30 + 5 \times 20^2 = 30 + 5 \times 400 = 30 + 2000 = 2030 \, \text{m}$

Thus, the position vector at $t = 20$s is: $\mathbf{r}(20) = 20050 \, \mathbf{i} + 2030 \, \mathbf{j} \, \text{m}$

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Do you want details or have any questions about this solution?
Here are some related questions for further exploration:

1. How would the path change if the acceleration was in both x and y directions?
2. How does the velocity vector change if the acceleration is turned off after 10 seconds?
3. What is the significance of the initial velocity vector in determining the path shape?
4. How would the equations change if there was an initial velocity component in the y-direction?
5. How do you find the total distance traveled by the spaceship by time $t = 20$ seconds?

Tip: Remember, when dealing with vector quantities, always break them down into their components to simplify calculations.