An object has an acceleration of 12 m/s2. What is the position of the object at 3
seconds if the velocity at time zero is 3 m/s?


To determine the position of the object at \( t = 3 \) seconds, we can use the kinematic equation for motion under constant acceleration:

\[
x(t) = x_0 + v_0 t + \frac{1}{2} a t^2
\]

Where:
- \( x(t) \) is the position at time \( t \),
- \( x_0 \) is the initial position (assumed to be 0 if not provided),
- \( v_0 \) is the initial velocity (3 m/s),
- \( a \) is the acceleration (12 m/s²),
- \( t \) is the time (3 seconds).

Given:
- \( x_0 = 0 \) (assuming the initial position is zero),
- \( v_0 = 3 \) m/s,
- \( a = 12 \) m/s²,
- \( t = 3 \) seconds.

Let's calculate the position:

\[
x(3) = 0 + 3 \times 3 + \frac{1}{2} \times 12 \times (3)^2
\]
\[
x(3) = 9 + \frac{1}{2} \times 12 \times 9
\]
\[
x(3) = 9 + 54 = 63 \, \text{meters}
\]

The position of the object at 3 seconds is **63 meters**.

Would you like any further explanation or details? Here are some related questions to explore:

1. What would the position be if the initial velocity were 0 m/s?
2. How would the position change if the acceleration was doubled?
3. What would the velocity of the object be at 3 seconds?
4. How long would it take for the object to reach a velocity of 30 m/s?
5. How would you calculate the displacement of the object over 5 seconds?

**Tip:** Kinematic equations are very useful for solving motion problems involving constant acceleration.

An object has an acceleration of 12 m/s². What is the position of the object at 3 seconds if the velocity at time zero is 3 m/s?

Learn how to calculate the position of an object after 3 seconds with an initial velocity of 3 m/s and an acceleration of 12 m/s² using kinematic equations. This solution demonstrates the use of the constant acceleration formula and is suitable for high school students in Grades 9-12.

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