A driver of a car traveling at 10 m/s presses the accelerator of his car causing a uniform acceleration of 2.0 m/s2.  The driver reaches a final speed of 15 m/s.  How far has the car moved during the acceleration period?

To determine the distance the car has moved during the acceleration period, we can use one of the kinematic equations that relates initial velocity (\(v_i\)), final velocity (\(v_f\)), acceleration (\(a\)), and displacement (\(d\)):

\[
v_f^2 = v_i^2 + 2ad
\]

Here:
- \(v_i = 10 \, \text{m/s}\) (initial velocity)
- \(v_f = 15 \, \text{m/s}\) (final velocity)
- \(a = 2.0 \, \text{m/s}^2\) (acceleration)
- \(d = ?\) (displacement, which we need to find)

We can rearrange the equation to solve for displacement \(d\):

\[
d = \frac{v_f^2 - v_i^2}{2a}
\]

Substituting the given values:

\[
d = \frac{(15 \, \text{m/s})^2 - (10 \, \text{m/s})^2}{2 \times 2.0 \, \text{m/s}^2}
\]

\[
d = \frac{225 \, \text{m}^2/\text{s}^2 - 100 \, \text{m}^2/\text{s}^2}{4 \, \text{m/s}^2}
\]

\[
d = \frac{125 \, \text{m}^2/\text{s}^2}{4 \, \text{m/s}^2}
\]

\[
d = 31.25 \, \text{m}
\]

So, the car has moved **31.25 meters** during the acceleration period.

Would you like more details or have any questions? 

Here are some related questions you might find interesting:
1. What is the time taken for the car to reach the final speed?
2. How would the displacement change if the initial speed was doubled?
3. What would be the displacement if the acceleration was halved?
4. How does the displacement change if the final speed was 20 m/s instead?
5. Can we determine the car's displacement using a different kinematic equation?

**Tip:** Understanding the kinematic equations and how they interrelate can greatly help in solving various motion-related problems effectively.

Learn how to calculate the distance moved by a car during acceleration using kinematic equations. This problem involves initial velocity, final velocity, acceleration, and displacement. Suitable for high school physics students.

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