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 }

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 }

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

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 }

Answer Each Question Appropriately dsdt=9.8t−3\frac { \mathrm { ds } } { \mathrm { dt } } = 9.8 \mathrm { t } - 3dtds​=9.8t−3

Multiple Choice