Specialist2025 Exam 2 Section A · Q17CAS active1 mark

Velocity from acceleration vector

The acceleration vector of a particle that starts from rest is given by

$\tvec{a}(t) = 4\cos(2t)\tvec{i} + 10\sin(2t)\tvec{j} - 6e^{-2t}\tvec{k}$ , where $t \geq 0$ .

The velocity vector of the particle, $v(t)$ , is given by

1 mark

\tvec{v}(t) = 2\sin(2t)\tvec{i} - 5\cos(2t)\tvec{j} + 3e^{-2t}\tvec{k}

\tvec{v}(t) = 2\sin(2t)\tvec{i} - 5\left(\cos(2t) + 1\right)\tvec{j} + 3(e^{-2t} + 1)\tvec{k}

\tvec{v}(t) = 2\sin(2t)\tvec{i} - 5\left(\cos(2t) - 1\right)\tvec{j} + 3(e^{-2t} - 1)\tvec{k}

\tvec{v}(t) = -8\sin(2t)\tvec{i} + 20\cos(2t)\tvec{j} + 12e^{-2t}\tvec{k}

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