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Find the tangential and normal components of the acceleration vector.

r(t) = cos(t)i + sin(t)j + tk at = an =

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Final answer:

To find the tangential and normal components of the acceleration vector, we can calculate the tangential acceleration (at) and the centripetal acceleration (ac). The tangential component of the acceleration vector is parallel to the velocity vector, while the normal component is perpendicular to the velocity vector.

Step-by-step explanation:

In order to find the tangential and normal components of the acceleration vector, we need to determine the tangential acceleration (at) and the centripetal acceleration (ac).

The tangential acceleration (at) represents the rate at which the magnitude of the velocity vector changes and is given by the formula at = d|v|/dt, where |v| is the magnitude of the velocity vector.

The centripetal acceleration (ac) represents the acceleration towards the center of the circular path and is given by the formula ac = v^2/r, where v is the magnitude of the velocity vector and r is the radius of the circular path.

Using the given information: a(t) = 5.0i + 2.0j – 6.0k m/s², we can now calculate the tangential and normal components of the acceleration vector.

Tangential component (at): The tangential component of the acceleration vector is parallel to the velocity vector. So, we only need to consider the tangential component of the given acceleration vector. In this case, at = 5.0i + 2.0j – 6.0k m/s².

Normal component (an): The normal component of the acceleration vector is perpendicular to the velocity vector. Since the tangential and normal components of the acceleration vector are perpendicular, we can find the normal component by subtracting the tangential component from the total acceleration. In this case, an = a(t) - a(c) = 5.0i + 2.0j – 6.0k m/s² - (0.0i + 0.0j + 24.8k m/s²) = 5.0i + 2.0j - 30.8k m/s².

User Misam
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Final answer:

To find the tangential and centripetal components of the acceleration, differentiate the position vector twice, project the resultant acceleration onto the velocity for the tangential component, and use Pythagoras' theorem for the centripetal component.

Step-by-step explanation:

To find the tangential and normal components (often called centripetal) of the acceleration vector, we need to differentiate the position vector r(t) with respect to time to find the velocity, and then differentiate the velocity to find acceleration.

The given position vector is r(t) = cos(t)i + sin(t)j + tk. The velocity vector v(t) is the first derivative of r(t), which gives v(t) = -sin(t)i + cos(t)j + k. The acceleration vector a(t) is the second derivative of r(t), which provides a(t) = -cos(t)i - sin(t)j. Notice that the k component disappears because the derivative of a constant is zero.

The tangential acceleration at is tangent to the path, and the normal or centripetal acceleration ac points towards the center of the circular path. Since the velocity vector is tangent to the path, we can find the tangential acceleration by projecting the acceleration vector onto the velocity vector using a dot product.

To find the magnitude of the tangential acceleration, we calculate at = a(t) \cdot (v(t)/|v(t)|). The centripetal acceleration can then be found using Pythagoras' theorem, given the total acceleration and the tangential acceleration: ac = √(|a(t)|^2 - at^2).

The total linear acceleration is the sum of the tangential and centripetal acceleration vectors and is given by a = ac + at. This is because the total acceleration a always points at an angle between ac and at, which are perpendicular to each other.

User Sony Packman
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