An iceboat on a frozen lake has acceleration . At $t=4s$, the iceboat's velocity is . What is the iceboat's displacement between $t=2s$ and $t=10s$?

• A.
• B.
• C.

Answer: (C)

Verified step by step guidance

1. Identify the given acceleration vector as a function of time: \(\mathbf{a}(t) = 2t \hat{\imath} + \left(1 - \frac{4}{t^3}\right) \hat{\jmath}\).
2. Recognize that velocity is the integral of acceleration with respect to time. To find the velocity function \(\mathbf{v}(t)\), integrate each component of \(\mathbf{a}(t)\) separately:
3. \[v_x(t) = \int 2t \, dt, \quad v_y(t) = \int \left(1 - \frac{4}{t^3}\right) dt.\]
4. Use the initial condition given at \(t=4\,s\), where \(\mathbf{v}(4) = 8 \hat{\imath} + 4.125 \hat{\jmath}\), to solve for the constants of integration in both \(v_x(t)\) and \(v_y(t)\).
5. Once \(\mathbf{v}(t)\) is known, find the displacement \(\Delta \mathbf{r}\) between \(t=2\,s\) and \(t=10\,s\) by integrating the velocity function over this interval:
6. \[\Delta \mathbf{r} = \int_{2}^{10} \mathbf{v}(t) \, dt.\]