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Question Number 37860 by Rio Mike last updated on 18/Jun/18

The distance moved by a particle  in t seconds is given by  s= t^3  + 3t + 1 where s is in   metres.Find the velocity and   Asseleration after 3 seconds.    Show all steps with short statements  on how the answers are gotten.

$${The}\:{distance}\:{moved}\:{by}\:{a}\:{particle} \\ $$$${in}\:{t}\:{seconds}\:{is}\:{given}\:{by} \\ $$$${s}=\:{t}^{\mathrm{3}} \:+\:\mathrm{3}{t}\:+\:\mathrm{1}\:{where}\:{s}\:{is}\:{in}\: \\ $$$${metres}.{Find}\:{the}\:{velocity}\:{and}\: \\ $$$${Asseleration}\:{after}\:\mathrm{3}\:{seconds}. \\ $$$$ \\ $$$${Show}\:{all}\:{steps}\:{with}\:{short}\:{statements} \\ $$$${on}\:{how}\:{the}\:{answers}\:{are}\:{gotten}. \\ $$

Answered by MrW3 last updated on 18/Jun/18

v(t)=(ds/dt)=3t^2 +3  a(t)=(dv/dt)=6t  at t=3 sec:  v=3×3^2 +3=30 m/s  a=6×3=18 m/s^2

$${v}\left({t}\right)=\frac{{ds}}{{dt}}=\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3} \\ $$$${a}\left({t}\right)=\frac{{dv}}{{dt}}=\mathrm{6}{t} \\ $$$${at}\:{t}=\mathrm{3}\:{sec}: \\ $$$${v}=\mathrm{3}×\mathrm{3}^{\mathrm{2}} +\mathrm{3}=\mathrm{30}\:{m}/{s} \\ $$$${a}=\mathrm{6}×\mathrm{3}=\mathrm{18}\:{m}/{s}^{\mathrm{2}} \\ $$

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