2.	An airplane accelerated uniformly from rest at the rate of 5.0 m/s2 South for 
14 s. What final velocity did it attain? 


We can solve this problem using the equation of motion for uniformly accelerated motion:

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

Where:
- \( v_f \) is the final velocity,
- \( v_i \) is the initial velocity (which is 0 m/s, since the airplane starts from rest),
- \( a \) is the acceleration (given as 5.0 m/s² South),
- \( t \) is the time (14 seconds).

Substituting the given values:

\[
v_f = 0 + (5.0 \, \text{m/s}^2) \cdot (14 \, \text{s})
\]
\[
v_f = 70 \, \text{m/s} \, \text{South}
\]

So, the final velocity attained by the airplane is **70 m/s South**.

Would you like more details or have any other questions?

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Here are 5 related questions:
1. How would the final velocity change if the acceleration were doubled?
2. What distance does the airplane cover during these 14 seconds?
3. How much time would it take to reach a velocity of 100 m/s under the same acceleration?
4. What is the difference between uniform and non-uniform acceleration?
5. How would the velocity be different if the airplane had started with an initial velocity of 20 m/s?

**Tip:** Always keep track of the direction in vector quantities like velocity and acceleration to avoid confusion.

An airplane accelerated uniformly from rest at the rate of 5.0 m/s² South for 14 s. What final velocity did it attain?

Calculate the final velocity of an airplane that starts from rest and accelerates uniformly at 5.0 m/s² South for 14 seconds. This physics problem involves kinematics and the use of the equation of motion. Suitable for students in Grades 9-12.

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