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

Question 149

Multiple Choice

Solve the problem.
-A particle moves on a coordinate line with acceleration a=d2s/dt2=(5/t) +9ta = d ^ { 2 } s / d t ^ { 2 } = ( 5 / \sqrt { t } ) + 9 \sqrt { t } , subject to the conditions that ds/dt=3\mathrm { ds } / \mathrm { dt } = 3 and s=1\mathrm { s } = 1 when t=1\mathrm { t } = 1 . Find the velocity v=ds/dt\mathrm { v } = \mathrm { ds } / \mathrm { dt } in terms of t\mathrm { t } and the position ss in terms of tt .


A) v=10t3+6t313;s=203t5+125t313t+7415v = 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 }
B) v=10t6t3+13;s=203t3125t5+13t+7415v = 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 }
C) v=10t+6t313;s=203t3+125t513t+7415v = 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 }
D) v=203t3+125t513t+7415;s=10t+6t313v = \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

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