The velocity of a plane as a function of time is given by . If the plane started from a point at $t=0$, what is it's position at $t=12s$?

• A.
• B.
• C.

Answer: (A)

Verified step by step guidance

1. Identify the given velocity vector as a function of time: \(\mathbf{v}(t) = 0.5t \hat{\imath} - 2 \cos(t) \hat{\jmath}\), and the initial position vector at \(t=0\) as \(\mathbf{r}(0) = 5.0 \hat{\imath}\).
2. Recall that position is the integral of velocity with respect to time, so express the position vector as \(\mathbf{r}(t) = \mathbf{r}(0) + \int_0^t \mathbf{v}(t') \, d t'\).
3. Set up the integral for each component separately: for the \(\hat{\imath}\) component, integrate \$0.5 t'$ from 0 to 12; for the \(\hat{\jmath}\) component, integrate \(-2 \cos(t')\) from 0 to 12.
4. Perform the integrations: the integral of \$0.5 t'\( with respect to \)t'\( is \)0.25 t'^2\(, and the integral of \(-2 \cos(t')\) with respect to \)t'$ is \(-2 \sin(t')\). Evaluate both definite integrals at the limits 0 and 12.
5. Add the initial position vector components to the results of the integrals to find the final position vector \(\mathbf{r}(12)\), combining the \(\hat{\imath}\) and \(\hat{\jmath}\) components accordingly.