Question

Find the velocity function v(t) of the car. Show that v is tangent to the circular curve. This means that, without a force to keep the car on the curve, the car will shoot off of it.

a) The velocity function v(t) of the car is:
b) v(t) = Rω(-sin(ωt), cos(ωt), 0)
c) v(t) = Rω(sin(ωt), -cos(ωt), 0)
d) v(t) = Rω(cos(ωt), sin(ωt), 0)

Answer 1

The correct velocity function is
`\( v(t)` =
`R\omega(-\sin(\omega t), \cos(\omega t), 0) \)`. It is tangent to the circular curve, signifying the car's motion without external force would cause it to leave the curve.

Step-by-step explanation:

Let's analyze the given velocity function and determine whether it is tangent to the circular curve.

The velocity function provided is
`\( v(t) = R\omega \begin{bmatrix} -\sin(\omega t) \\ \cos(\omega t) \\ 0 \end{bmatrix} \).`

Here,
`\( R \)` represents the radius of the circular curve
`, \( \omega \)`is the angular velocity, and
`, \( \omega \)` is time.

To show that
`\( v(t) \)` is tangent to the circular curve, we need to find the position vector
`\( r(t) \)` of a point on the curve and then take the derivative of
`\( r(t) \)`to obtain the velocity vector
`\( v(t) \)`. If
`\( v(t) \)` is parallel to
`\( r(t) \)`, it indicates that
`\( v(t) \)`is tangent to the curve.

Assuming the circular curve is in the xy-plane, the position vector can be expressed as
`\( r(t) = R \begin{bmatrix} \cos(\omega t) \\ \sin(\omega t) \\ 0 \end{bmatrix} \).`

Now, let's find the derivative of
`\( r(t) \)` with respect to
`\( t \)`to obtain the velocity vector:

`\[ (dr)/(dt) = R\omega \begin{bmatrix} -\sin(\omega t) \\ \cos(\omega t) \\ 0 \end{bmatrix} \].`

`Comparing this with the given \( v(t) \), we can see that \( v(t) \) is indeed parallel to \( r(t) \), which means \( v(t) \) is tangent to the circular curve.`

`Therefore, the correct answer is option \( \mathbf{a) \ v(t) = R\omega \begin{bmatrix} -\sin(\omega t) \\ \cos(\omega t) \\ 0 \end{bmatrix}} \).`