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The Position Vector of a Particle Is R(t) t=π8t = \frac { \pi } { 8 }

Question 126

Multiple Choice

The position vector of a particle is r(t) . Find the requested vector.
-The acceleration at t=π8t = \frac { \pi } { 8 } for r(t) =(10t3t3) i+2tan(2t) j+e5tk\mathbf { r } ( \mathrm { t } ) = \left( 10 \mathrm { t } - 3 \mathrm { t } ^ { 3 } \right) \mathbf { i } + 2 \tan ( 2 \mathrm { t } ) \mathbf { j } + \mathrm { e } ^ { 5 } \mathrm { t } \mathbf { k }


A) a(π8) =94πii32j+25ka \left( \frac { \pi } { 8 } \right) = - \frac { 9 } { 4 } \pi i \mathbf { i } - 32 \mathbf { j } + 25 \mathbf { k }
B) a(π8) =94πi+32j+25e5/8πka \left( \frac { \pi } { 8 } \right) = \frac { 9 } { 4 } \pi \mathbf { i } + 32 \mathbf { j } + 25 \mathrm { e } ^ { 5 / 8 } \pi \mathbf { k }
C) a(π8) =94πi+32j+25e5/8πka \left( \frac { \pi } { 8 } \right) = - \frac { 9 } { 4 } \pi \mathrm { i } + 32 \mathrm { j } + 25 \mathrm { e } ^ { 5 / 8 \pi } \mathbf { k }
D) a(π8) =94πi32j+25e5/8πka \left( \frac { \pi } { 8 } \right) = - \frac { 9 } { 4 } \pi i - 32 j + 25 e ^ { 5 / 8 } \pi k

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