1. A car begins at rest at a stoplight. After the light turn green, the car begins to accelerate
5 m/s2 over the span of 3.6 seconds. What is the car’s displacement over this period of
time?

To solve this problem, we can use one of the kinematic equations for motion under constant acceleration:

\[
s = v_0 t + \frac{1}{2} a t^2
\]

where:
- \( s \) is the displacement (what we're solving for),
- \( v_0 \) is the initial velocity (0 m/s, since the car starts from rest),
- \( a \) is the acceleration (5 m/s\(^2\)),
- \( t \) is the time (3.6 seconds).

Since \( v_0 = 0 \), the equation simplifies to:

\[
s = \frac{1}{2} a t^2
\]

Now, substitute the given values into the equation:

\[
s = \frac{1}{2} \times 5 \, \text{m/s}^2 \times (3.6 \, \text{s})^2
\]

\[
s = \frac{1}{2} \times 5 \times 12.96 = 2.5 \times 12.96 = 32.4 \, \text{m}
\]

So, the car's displacement over the period of 3.6 seconds is **32.4 meters**.

Would you like more details on the steps involved, or do you have any questions?

Here are 5 related questions:
1. What would be the final velocity of the car at the end of the 3.6 seconds?
2. How much time would it take for the car to reach a displacement of 100 meters under the same acceleration?
3. If the car maintained the final velocity from the acceleration phase, how far would it travel in the next 5 seconds?
4. How would the displacement change if the acceleration were halved?
5. What is the car's average velocity during the acceleration phase?

**Tip:** In problems involving acceleration from rest, the displacement is directly proportional to the square of the time.

A car begins at rest at a stoplight. After the light turns green, the car begins to accelerate 5 m/s² over the span of 3.6 seconds. What is the car’s displacement over this period of time?

Learn how to calculate the displacement of a car accelerating at 5 m/s² over 3.6 seconds from rest using the kinematic equation s = ½at². This problem explains key physics concepts of motion and is ideal for students in Grades 9-11.

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