\(\begin{pmatrix} -14 \\ 15 \end{pmatrix} ms^{-1}\)
\(\begin{pmatrix} -2 \\ 1 \end{pmatrix} ms^{-1}\)
\(\begin{pmatrix} 4 \\ -9 \end{pmatrix} ms^{-1}\)
\(\begin{pmatrix} 14 \\ -9 \end{pmatrix} ms^{-1}\)
Correct answer is C
\(u = \begin{pmatrix} -5 \\ 3 \end{pmatrix} ms^{-1}\)
\(a = \begin{pmatrix} 3 \\ -4 \end{pmatrix} ms^{-2}; t = 3 secs\)
\(v = u + at \implies v = \begin{pmatrix} -5 \\ 3 \end{pmatrix} + \begin{pmatrix} 3 \\ -4 \end{pmatrix} \times 3\)
= \(\begin{pmatrix} -5 \\ 3 \end{pmatrix} + \begin{pmatrix} 9 \\ -12 \end{pmatrix} = \begin{pmatrix} 4 \\ -9 \end{pmatrix} ms^{-1}\)\)