Dirive an expression for the car’s maximum speed wthin the interval 0<=t<=3p/2q

To derive an expression for the car's maximum speed within the time interval \( 0 \leq t \leq \frac{3\pi}{2q} \), we need to first define what speed refers to in this context. 

### Step-by-Step Process:

1. **Speed and Velocity:**
   Speed is the magnitude of the velocity function. If we have a position function \( x(t) \), the velocity \( v(t) \) is the first derivative of the position function with respect to time:
   \[
   v(t) = \frac{dx(t)}{dt}
   \]
   The speed is the absolute value of velocity:
   \[
   \text{Speed} = |v(t)|
   \]

2. **Form of the Position Function:**
   Assume the position function \( x(t) \) is periodic, because the interval involves \( \pi \) and \( q \), which often implies trigonometric functions like sine or cosine. A common form for such a position function is:
   \[
   x(t) = A \sin(qt)
   \]
   where \( A \) is the amplitude and \( q \) is related to the frequency.

3. **Find the Velocity Function:**
   The velocity is the derivative of \( x(t) \):
   \[
   v(t) = \frac{d}{dt} [A \sin(qt)] = A q \cos(qt)
   \]

4. **Find the Critical Points:**
   To find when the speed is maximum, we examine the velocity function and set its derivative equal to zero. The derivative of velocity is:
   \[
   a(t) = \frac{dv(t)}{dt} = -A q^2 \sin(qt)
   \]
   Setting \( a(t) = 0 \), we get

Derive an expression for the car’s maximum speed within the interval 0 <= t <= 3π/2q

Explore how to derive the maximum speed of a car within the time interval 0 <= t <= 3π/2q using calculus and trigonometric principles. The step-by-step solution covers velocity, speed, and the critical points of a sinusoidal function. Suitable for advanced high school or college calculus students.

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