The position vector of a particle is r(t). Find the requested vector.
-The acceleration at $t = 0$ for $r ( t ) = t ^ { 2 } i + \left( 8 t ^ { 3 } - 10 \right) j + \sqrt { 4 - 3 t } k$
A) $\mathbf { a } ( 0 ) = 2 \mathbf { i } - \frac { 3 } { 32 } \mathbf { k }$
B) $\mathbf { a } ( 0 ) = 2 \mathbf { i } + \frac { 9 } { 32 } \mathbf { k }$
C) $\mathbf { a } ( 0 ) = 2 \mathbf { i } - \frac { 9 } { 8 } \mathbf { k }$
D) $\mathbf { a } ( 0 ) = 2 \mathbf { i } - \frac { 9 } { 32 } \mathbf { k }$

Question

Quizplus · Accepted Answer

Acceleration is the second derivative of position with respect to time. Thus, we need to find the second derivative of r(t) and evaluate it at t=0.
r(t) = t^2i + (8t^3-10)j + sqrt(4-3t)k
r'(t) = 2ti + (24t^2)j + (-3/2)(4-3t)^(-1/2)k
r''(t) = 2i + (48t)j + (9/4)(4-3t)^(-3/2)
r''(0) = 2i - (9/32)k
Therefore, the option (D) is the correct answer.

The Position Vector of a Particle Is R(t) t=0t = 0t=0

The Position Vector of a Particle Is R(t) $t = 0$