Solve the problem.
-A particle moves on a coordinate line with acceleration $a = d ^ { 2 } s / d t ^ { 2 } = ( 5 / \sqrt { t } ) + 9 \sqrt { t }$, subject to the conditions that $\mathrm { ds } / \mathrm { dt } = 3$ and $\mathrm { s } = 1$ when $\mathrm { t } = 1$. Find the velocity $\mathrm { v } = \mathrm { ds } / \mathrm { dt }$ in terms of $\mathrm { t }$ and the position $s$ in terms of $t$.
A) $v = 10 \sqrt { t } + 6 \sqrt [ 3 ] { t } - 13 ; s = \frac { 20 } { 3 } \sqrt [ 3 ] { t } + \frac { 12 } { 5 } \sqrt [ 5 ] { t } - 13 t + \frac { 74 } { 15 }$
B) $v = 10 \sqrt [ 3 ] { t } + 6 \sqrt [ 3 ] { t } - 13 ; s = \frac { 20 } { 3 } \sqrt [ 5 ] { t } + \frac { 12 } { 5 } \sqrt [ 3 ] { t } - 13 t + \frac { 74 } { 15 }$
C) $v = \frac { 20 } { 3 } \sqrt [ 3 ] { t } + \frac { 12 } { 5 } \sqrt [ 5 ] { t } - 13 t + \frac { 74 } { 15 } ; s = 10 \sqrt { t } + 6 \sqrt [ 3 ] { t } - 13$
D) $v = 10 \sqrt { t } - 6 \sqrt [ 3 ] { t } + 13 ; s = \frac { 20 } { 3 } \sqrt [ 3 ] { t } - \frac { 12 } { 5 } \sqrt [ 5 ] { t } + 13 t + \frac { 74 } { 15 }$

Question

Quizplus · Accepted Answer

To find the velocity $v = ds/dt$, integrate the acceleration $a = d^2s/dt^2 = (5/\sqrt{t}) + 9\sqrt{t}$ with respect to $t$. Integrating gives $v = 10\sqrt{t} + 6t^{2/3} + C$. Using the initial condition $ds/dt = 3$ when $t = 1$, we find $C = -13$, so $v = 10\sqrt{t} + 6t^{2/3} - 13$. To find the position $s$, integrate $v$ with respect to $t$, yielding $s = \frac{20}{3}t^{3/2} + \frac{12}{5}t^{5/3} - 13t + C'$. Using the initial condition $s = 1$ when $t = 1$, we find $C' = \frac{74}{15}$, so $s = \frac{20}{3}t^{3/2} + \frac{12}{5}t^{5/3} - 13t + \frac{74}{15}$.

Solve the Problem a=d2s/dt2=(5/t)+9ta = d ^ { 2 } s / d t ^ { 2 } = ( 5 / \sqrt { t } ) + 9 \sqrt { t }a=d2s/dt2=(5/t​)+9t​

Solve the Problem $a = d ^ { 2 } s / d t ^ { 2 } = ( 5 / \sqrt { t } ) + 9 \sqrt { t }$