The image you've uploaded contains a physics problem involving motion with a velocity function. Let's break down the problem and proceed step by step to find the answers.

The question states:

a) At the time when the object starts moving, its velocity is zero. What is the velocity at that time, as described by the function $v(t) = 2\sin(t)$?

b) Find the velocity of the object at the time $t = \frac{\pi}{2}$ seconds.

c) Find the distance the object travels from $t = 0$ to $t = \pi$ seconds.

d) Find the distance the object travels from $t = \frac{\pi}{2}$ to $t = \frac{3\pi}{4}$ seconds.

I will now begin solving these one by one.

For a):

• Since velocity $v(t)$ is the derivative of position $x(t)$, the function for velocity is given as $v(t) = 2\sin(t)$.
• For the initial condition, the velocity at $t = 0$ is $v(0) = 2\sin(0) = 0$, which matches the statement that the object starts from rest.

For b):

• To find the velocity at $t = \frac{\pi}{2}$, we substitute this value into the velocity equation:
  $v\left( \frac{\pi}{2} \right) = 2\sin\left( \frac{\pi}{2} \right) = 2 \times 1 = 2 \, \text{m/s}$

For c):

• To find the distance traveled, we must integrate the velocity function. The distance is the integral of $v(t)$ from $t = 0$ to $t = \pi$: $x(t) = \int v(t) \, dt = \int 2\sin(t) \, dt = -2\cos(t) + C$ The constant $C$ can be determined by knowing that at $t = 0$, the object starts at the origin, so $x(0) = 0$: $x(0) = -2\cos(0) + C = 0 \quad \Rightarrow \quad C = 2$ Therefore, the position function is: $x(t) = -2\cos(t) + 2$ Now, to find the distance traveled from $t = 0$ to $t = \pi$: = \left(-2(-1) + 2\right) - \left(-2(1) + 2\right) = 4 - 0 = 4 \, \text{meters}$$

For d):

• To find the distance from $t = \frac{\pi}{2}$ to $t = \frac{3\pi}{4}$, we use the same position function. The distance is: $\text{Distance} = x\left( \frac{3\pi}{4} \right) - x\left( \frac{\pi}{2} \right)$ First, compute $x\left( \frac{\pi}{2} \right)$ and $x\left( \frac{3\pi}{4} \right)$: $x\left( \frac{\pi}{2} \right) = -2\cos\left( \frac{\pi}{2} \right) + 2 = -2(0) + 2 = 2$ $x\left( \frac{3\pi}{4} \right) = -2\cos\left( \frac{3\pi}{4} \right) + 2 = -2\left(-\frac{1}{\sqrt{2}}\right) + 2 = 2\left(\frac{1}{\sqrt{2}}\right) + 2$ Calculate the distance: $\text{Distance} = \left( 2\left(\frac{1}{\sqrt{2}}\right) + 2 \right) - 2 = 2\left(\frac{1}{\sqrt{2}}\right) = \text{approximately} \, 2 \, \text{m}$

I hope that clarifies everything! Would you like further details on any of these steps? Let me know if you have any questions or need more help!

Here are some related questions:

1. How can you calculate the velocity and position for more complex trigonometric functions?
2. What happens to the motion if the acceleration function is non-sinusoidal?
3. How do we handle negative velocities in physics problems?
4. How do you calculate the maximum and minimum velocities of an object from its velocity function?
5. What role does the integration constant play in motion problems?

Tip: When dealing with trigonometric motion functions, always remember to apply the correct boundary conditions (like initial positions or velocities) to find the integration constants.