The position of an object is given by $\overrightarrow { \mathbf { r } } = \left( - 4.00 t \hat { \mathbf { i } } + 6.00 t ^ { 3 } \hat { \mathbf { j } } \right) \mathrm { m }$ where t is in seconds.At t = 2.0 s,what is the magnitude of the particle's acceleration? A)0 m/s² B)2.0 m/s² C)17 m/s² D)36 m/s² E)72 m/s²

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Quizplus · Accepted Answer

The acceleration is the second derivative of the position vector. The position vector is $\overrightarrow { \mathbf { r } } = \left( - 4.00 t \hat { \mathbf { i } } + 6.00 t ^ { 3 } \hat { \mathbf { j } } \right) \mathrm { m }$. The velocity is $\overrightarrow { \mathbf { v } } = \frac{d}{dt}(-4.00t \hat{\mathbf{i}} + 6.00t^3 \hat{\mathbf{j}}) = -4.00 \hat{\mathbf{i}} + 18.00t^2 \hat{\mathbf{j}}$. The acceleration is $\overrightarrow { \mathbf { a } } = \frac{d}{dt}(-4.00 \hat{\mathbf{i}} + 18.00t^2 \hat{\mathbf{j}}) = 36.00t \hat{\mathbf{j}}$. At $t = 2.0$ s, $\overrightarrow { \mathbf { a } } = 36.00 \times 2.0 \hat{\mathbf{j}} = 72.00 \hat{\mathbf{j}}$ m/s². The magnitude is 72 m/s².

The Position of an Object Is Given By r→=(−4.00ti^+6.00t3j^)m\overrightarrow { \mathbf { r } } = \left( - 4.00 t \hat { \mathbf { i } } + 6.00 t ^ { 3 } \hat { \mathbf { j } } \right) \mathrm { m }r=(−4.00ti^+6.00t3j^​)m

The Position of an Object Is Given By $\overrightarrow { \mathbf { r } } = \left( - 4.00 t \hat { \mathbf { i } } + 6.00 t ^ { 3 } \hat { \mathbf { j } } \right) \mathrm { m }$