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Sagot :
To determine the velocity of a particle given its acceleration vector [tex]\(\vec{a}\)[/tex], we need to integrate the acceleration with respect to time. Here, the acceleration vector is provided as:
[tex]\[ \vec{a} = \cos(t) \vec{i} + \sin(t) \vec{j} + 3t \vec{k} \][/tex]
Given that the particle starts from rest, the initial velocity vector [tex]\(\vec{v}(0) = 0\)[/tex].
### Step-by-Step Solution:
1. Express the components of acceleration:
- [tex]\(a_x = \cos(t)\)[/tex]
- [tex]\(a_y = \sin(t)\)[/tex]
- [tex]\(a_z = 3t\)[/tex]
2. Integrate each component of the acceleration with respect to [tex]\(t\)[/tex] to find the components of the velocity:
- The [tex]\(x\)[/tex]-component of velocity:
[tex]\[ v_x(t) = \int \cos(t) \, dt \][/tex]
- The [tex]\(y\)[/tex]-component of velocity:
[tex]\[ v_y(t) = \int \sin(t) \, dt \][/tex]
- The [tex]\(z\)[/tex]-component of velocity:
[tex]\[ v_z(t) = \int 3t \, dt \][/tex]
3. Compute each integral:
- For [tex]\(v_x(t)\)[/tex]:
[tex]\[ v_x(t) = \int \cos(t) \, dt = \sin(t) + C_1 \][/tex]
- For [tex]\(v_y(t)\)[/tex]:
[tex]\[ v_y(t) = \int \sin(t) \, dt = -\cos(t) + C_2 \][/tex]
- For [tex]\(v_z(t)\)[/tex]:
[tex]\[ v_z(t) = \int 3t \, dt = \frac{3t^2}{2} + C_3 \][/tex]
4. Apply the initial condition ([tex]\(\vec{v}(0) = 0\)[/tex]) to find the constants of integration [tex]\(C_1\)[/tex], [tex]\(C_2\)[/tex], and [tex]\(C_3\)[/tex]:
- When [tex]\(t = 0\)[/tex]:
[tex]\[ 0 = \sin(0) + C_1 \implies C_1 = 0 \][/tex]
[tex]\[ 0 = -\cos(0) + C_2 \implies C_2 = 1 \][/tex]
[tex]\[ 0 = \frac{3 \cdot 0^2}{2} + C_3 \implies C_3 = 0 \][/tex]
5. Substitute the constants back into the velocity components:
- [tex]\(v_x(t) = \sin(t)\)[/tex]
- [tex]\(v_y(t) = -\cos(t) + 1\)[/tex]
- [tex]\(v_z(t) = \frac{3t^2}{2}\)[/tex]
6. Combine the components to form the velocity vector [tex]\(\vec{v}(t)\)[/tex]:
[tex]\[ \vec{v}(t) = \sin(t) \vec{i} + (-\cos(t) + 1) \vec{j} + \frac{3t^2}{2} \vec{k} \][/tex]
### Final Answer:
The velocity of the particle as a function of time [tex]\(t\)[/tex], given that it starts from rest, is:
[tex]\[ \boxed{\vec{v}(t) = \sin(t) \vec{i} + (-\cos(t) + 1) \vec{j} + \frac{3t^2}{2} \vec{k}} \][/tex]
[tex]\[ \vec{a} = \cos(t) \vec{i} + \sin(t) \vec{j} + 3t \vec{k} \][/tex]
Given that the particle starts from rest, the initial velocity vector [tex]\(\vec{v}(0) = 0\)[/tex].
### Step-by-Step Solution:
1. Express the components of acceleration:
- [tex]\(a_x = \cos(t)\)[/tex]
- [tex]\(a_y = \sin(t)\)[/tex]
- [tex]\(a_z = 3t\)[/tex]
2. Integrate each component of the acceleration with respect to [tex]\(t\)[/tex] to find the components of the velocity:
- The [tex]\(x\)[/tex]-component of velocity:
[tex]\[ v_x(t) = \int \cos(t) \, dt \][/tex]
- The [tex]\(y\)[/tex]-component of velocity:
[tex]\[ v_y(t) = \int \sin(t) \, dt \][/tex]
- The [tex]\(z\)[/tex]-component of velocity:
[tex]\[ v_z(t) = \int 3t \, dt \][/tex]
3. Compute each integral:
- For [tex]\(v_x(t)\)[/tex]:
[tex]\[ v_x(t) = \int \cos(t) \, dt = \sin(t) + C_1 \][/tex]
- For [tex]\(v_y(t)\)[/tex]:
[tex]\[ v_y(t) = \int \sin(t) \, dt = -\cos(t) + C_2 \][/tex]
- For [tex]\(v_z(t)\)[/tex]:
[tex]\[ v_z(t) = \int 3t \, dt = \frac{3t^2}{2} + C_3 \][/tex]
4. Apply the initial condition ([tex]\(\vec{v}(0) = 0\)[/tex]) to find the constants of integration [tex]\(C_1\)[/tex], [tex]\(C_2\)[/tex], and [tex]\(C_3\)[/tex]:
- When [tex]\(t = 0\)[/tex]:
[tex]\[ 0 = \sin(0) + C_1 \implies C_1 = 0 \][/tex]
[tex]\[ 0 = -\cos(0) + C_2 \implies C_2 = 1 \][/tex]
[tex]\[ 0 = \frac{3 \cdot 0^2}{2} + C_3 \implies C_3 = 0 \][/tex]
5. Substitute the constants back into the velocity components:
- [tex]\(v_x(t) = \sin(t)\)[/tex]
- [tex]\(v_y(t) = -\cos(t) + 1\)[/tex]
- [tex]\(v_z(t) = \frac{3t^2}{2}\)[/tex]
6. Combine the components to form the velocity vector [tex]\(\vec{v}(t)\)[/tex]:
[tex]\[ \vec{v}(t) = \sin(t) \vec{i} + (-\cos(t) + 1) \vec{j} + \frac{3t^2}{2} \vec{k} \][/tex]
### Final Answer:
The velocity of the particle as a function of time [tex]\(t\)[/tex], given that it starts from rest, is:
[tex]\[ \boxed{\vec{v}(t) = \sin(t) \vec{i} + (-\cos(t) + 1) \vec{j} + \frac{3t^2}{2} \vec{k}} \][/tex]
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