To find the magnitude of the spaceship's velocity $v$ after $t = 5.94 \, \text{s}$, we will calculate the velocity components in each direction and then determine the magnitude of the velocity vector.

Given:

• Initial velocity: $\vec{v}_0 = (39.9 \, \text{m/s}) \hat{i}$
• Acceleration: $\vec{a} = (3.52 \, \text{m/s}^2) \hat{i} + (4.27 \, \text{m/s}^2) \hat{j}$
• Time: $t = 5.94 \, \text{s}$

Step 1: Find the velocity components after time $t$

Using the equation for velocity: $\vec{v} = \vec{v}_0 + \vec{a} t$ The components are calculated as:

1. In the $\hat{i}$-direction: $v_x = v_{0x} + a_x t = 39.9 + (3.52)(5.94)$ $v_x = 39.9 + 20.93 = 60.83 \, \text{m/s}$

2. In the $\hat{j}$-direction: $v_y = v_{0y} + a_y t = 0 + (4.27)(5.94)$ $v_y = 25.37 \, \text{m/s}$

Step 2: Calculate the magnitude of $\vec{v}$

The magnitude of the velocity $v$ is given by: $v = \sqrt{v_x^2 + v_y^2}$ Substitute the values: $v = \sqrt{(60.83)^2 + (25.37)^2}$ $v = \sqrt{3699.89 + 643.68} = \sqrt{4343.57}$ $v \approx 65.88 \, \text{m/s}$

Final Answer:

The spaceship's speed is approximately: $v \approx 65.88 \, \text{m/s}$

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Would you like a detailed explanation of any step or related concepts?

Related Questions:

1. What is the significance of calculating velocity components?
2. How would the calculation change if the acceleration in the $\hat{i}$-direction was negative?
3. What is the importance of vector magnitude in physics problems?
4. How can we find the direction of the velocity vector after 5.94 seconds?
5. How would the result change if the initial velocity had a $\hat{j}$-component?

Tip: Always resolve vectors into components for problems involving motion in two or three dimensions.