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The problem provides the equation for the displacement \( s \) of an object at time \( t \), given by:

\[
s = 10 \sin 2t
\]

where \( s \) represents the displacement in meters. We are tasked with finding:

a. The velocity of the object at \( t = \frac{\pi}{6} \).

b. The acceleration of the object at \( t = \frac{3\pi}{4} \).

### Step-by-Step Solution:

1. **Velocity:**
   Velocity \( v(t) \) is the derivative of displacement \( s(t) \) with respect to time \( t \).

   \[
   v(t) = \frac{ds}{dt} = \frac{d}{dt} \left( 10 \sin 2t \right)
   \]

   Using the chain rule, the derivative of \( \sin(2t) \) is \( 2 \cos(2t) \). Therefore:

   \[
   v(t) = 10 \cdot 2 \cos 2t = 20 \cos 2t
   \]

   Now, we substitute \( t = \frac{\pi}{6} \) to find the velocity at that time:

   \[
   v\left( \frac{\pi}{6} \right) = 20 \cos \left( 2 \cdot \frac{\pi}{6} \right) = 20 \cos \left( \frac{\pi}{3} \right)
   \]

   Since \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \):

   \[
   v\left( \frac{\pi}{6} \right) = 20 \times \frac{1}{2} = 10 \, \text{m/s}
   \]

2. **Acceleration:**
   Acceleration \( a(t) \) is the derivative of velocity \( v(t) \) with respect to time \( t \).

   \[
   a(t) = \frac{dv}{dt} = \frac{d}{dt} \left( 20 \cos 2t \right)
   \]

   Again using the chain rule, the derivative of \( \cos(2t) \) is \( -2 \sin(2t) \):

   \[
   a(t) = 20 \cdot (-2) \sin 2t = -40 \sin 2t
   \]

   Now, we substitute \( t = \frac{3\pi}{4} \) to find the acceleration at that time:

   \[
   a\left( \frac{3\pi}{4} \right) = -40 \sin \left( 2 \cdot \frac{3\pi}{4} \right) = -40 \sin \left( \frac{3\pi}{2} \right)
   \]

   Since \( \sin \left( \frac{3\pi}{2} \right) = -1 \):

   \[
   a\left( \frac{3\pi}{4} \right) = -40 \times (-1) = 40 \, \text{m/s}^2
   \]

### Final Answers:
- **Velocity at \( t = \frac{\pi}{6} \):** \( 10 \, \text{m/s} \)
- **Acceleration at \( t = \frac{3\pi}{4} \):** \( 40 \, \text{m/s}^2 \)

---

Would you like further details or explanations? Here are some related questions to consider:

1. What would be the velocity at \( t = \frac{\pi}{3} \)?
2. How would the acceleration change if the equation was \( s = 5 \sin 2t \)?
3. Can you calculate the displacement when \( t = \frac{\pi}{4} \)?
4. What is the significance of the negative sign in acceleration?
5. How does changing the frequency of \( \sin(2t) \) affect the velocity and acceleration?

**Tip:** Always remember, the derivative of \( \sin(kx) \) is \( k \cos(kx) \), and the derivative of \( \cos(kx) \) is \( -k \sin(kx) \).

Perpindahan sebuah benda pada saat t detik diberikan oleh persamaan s = 10 sin 2t, dengan s adalah jarak yang dinyatakan dalam meter. Tentukan: a. Kecepatan benda tersebut pada saat t = pi/6 b. Percepatan benda tersebut pada saat t = 3pi/4.

Learn how to find the velocity and acceleration of an object when given the displacement s = 10 sin 2t. This step-by-step solution covers trigonometric derivatives and kinematic equations. Suitable for students in Grades 11-12.

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