superman is flying at 300m/s. he then accelerates at a rate of 29m/s^2 for 10 seconds. how fast is he now flying 

To calculate Superman's final speed after accelerating, we can use the following kinematic equation:

\[
v_f = v_i + a \cdot t
\]

Where:
- \( v_f \) is the final velocity (what we need to find),
- \( v_i = 300 \, \text{m/s} \) is the initial velocity,
- \( a = 29 \, \text{m/s}^2 \) is the acceleration,
- \( t = 10 \, \text{seconds} \) is the time.

Now plug the values into the equation:

\[
v_f = 300 \, \text{m/s} + (29 \, \text{m/s}^2 \cdot 10 \, \text{seconds})
\]

\[
v_f = 300 \, \text{m/s} + 290 \, \text{m/s}
\]

\[
v_f = 590 \, \text{m/s}
\]

So, after accelerating for 10 seconds, Superman is flying at **590 m/s**.

Would you like any details or have any questions? 

Here are five related questions:
1. What if the acceleration lasted for 20 seconds instead?
2. How far does Superman travel during the 10-second acceleration period?
3. What would his final speed be if he had accelerated at 50 m/s² for 10 seconds?
4. How long would it take for Superman to reach 1000 m/s if the acceleration rate remained 29 m/s²?
5. If he started at rest and accelerated for 10 seconds at 29 m/s², how fast would he be?

**Tip:** When solving problems involving acceleration, always check if the motion is uniform or involves changes in velocity over time.

Superman is flying at 300m/s. He then accelerates at a rate of 29m/s^2 for 10 seconds. How fast is he now flying?

Find out how fast Superman is flying after accelerating at 29m/s² for 10 seconds. This problem uses kinematic equations to calculate the final velocity, making it a great example of uniform acceleration and linear motion. Suitable for students in Grades 9-12.

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