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The given displacement equation is:

\[
s(t) = 15t^2 - 2t + 5 \, \text{m}
\]

### 1. Velocity at \( t = 1 \) second:
Velocity is the first derivative of displacement with respect to time \( t \). Therefore, we differentiate \( s(t) \) to get velocity:

\[
v(t) = \frac{ds(t)}{dt} = \frac{d}{dt} \left( 15t^2 - 2t + 5 \right) = 30t - 2
\]

Now, substitute \( t = 1 \):

\[
v(1) = 30(1) - 2 = 30 - 2 = 28 \, \text{m/s}
\]

### 2. Acceleration at \( t = 2 \) seconds:
Acceleration is the first derivative of velocity with respect to time, or the second derivative of displacement. So, we differentiate the velocity function:

\[
a(t) = \frac{dv(t)}{dt} = \frac{d}{dt} \left( 30t - 2 \right) = 30
\]

Since the acceleration is constant, it remains 30 m/s² at all times, including \( t = 2 \) seconds.

### Final answers:
1. Velocity at \( t = 1 \) second is \( 28 \, \text{m/s} \).
2. Acceleration at \( t = 2 \) seconds is \( 30 \, \text{m/s}^2 \).

Would you like more details, or have any further questions? 

Here are five related questions to consider:
1. How do you find displacement at any given time from the velocity?
2. What is the physical meaning of constant acceleration in this context?
3. How would the problem change if the displacement equation included a cubic term?
4. Can you determine the time at which the velocity is zero?
5. What would be the distance traveled in the first 2 seconds?

**Tip:** The derivative of a position function with respect to time always gives you the velocity, and the derivative of the velocity gives you the acceleration.

Displacement of a car is s(t) = 15t^2 - 2t + 5 meters. Find the velocity at t = 1 second and the acceleration at t = 2 seconds.

Learn how to calculate the velocity and acceleration of a car from its displacement function s(t) = 15t^2 - 2t + 5. Detailed solution involves using derivatives to find velocity at t = 1 second and acceleration at t = 2 seconds. Suitable for Grades 10-12 students.

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