Vector Valued Function Acceleration

Question

The position of a particle moving in the $$xy$$-plane is given by the vector $$\left\langle 4t^3, y(2t) \right\rangle$$, where $$y$$ is a twice-differentiable function. What is the acceleration vector of the particle at time $$t = \dfrac{1}{2}$$?

Accepted Answer

$$\displaystyle \left\langle 12, 4y''(1) \right\rangle$$

Answer

$$\displaystyle \left\langle 12, 2y''(1) \right\rangle$$

Answer

$$\displaystyle \left\langle 6, 4y''(1) \right\rangle$$

Answer

$$\displaystyle \left\langle 3, 2y''(1) \right\rangle$$

Vector Valued Function Acceleration

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