**An object starting from rest accelerates at a rate of 3.0 meters/seconds squared for 6.0 seconds. The velocity of the object at the end of this time is**

To calculate the velocity of the object at the end of the given time, we use the formula for velocity under constant acceleration:

\[
v = u + at
\]

Where:  
- \( v \) = final velocity  
- \( u \) = initial velocity (0, since the object starts from rest)  
- \( a \) = acceleration (\(3.0 \, \text{m/s}^2\))  
- \( t \) = time (\(6.0 \, \text{s}\))  

Substitute the given values into the formula:

\[
v = 0 + (3.0 \, \text{m/s}^2)(6.0 \, \text{s}) = 18.0 \, \text{m/s}
\]

### Final Answer:
The velocity of the object at the end of 6.0 seconds is **18.0 m/s**.

Would you like a detailed explanation of the steps? Let me know if you have any questions.  

### Follow-up Questions:
1. What would be the distance traveled by the object during this time?  
2. How would the velocity change if the acceleration were doubled?  
3. What is the kinetic energy of the object at the end of this time, assuming its mass is known?  
4. What would the velocity be if the object accelerated for 8.0 seconds instead of 6.0 seconds?  
5. How would the result change if the object started with an initial velocity of 5.0 m/s?  

### Tip:  
When solving kinematics problems, always check whether the object starts from rest or with an initial velocity (\(u\)). This affects how the equations are applied.

An object starting from rest accelerates at a rate of 3.0 meters/seconds squared for 6.0 seconds. The velocity of the object at the end of this time is

Learn how to calculate the velocity of an object starting from rest and accelerating at 3.0 m/s² for 6.0 seconds. This problem involves the basic kinematic equation of motion and is suitable for students in Grades 9-10.

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