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Question Number 13608 by chintan vasoya last updated on 21/May/17

if distance is given by x(t)=2t+5 then ,the  acceleration at 4s is............

$$\mathrm{if}\:\mathrm{distance}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{x}\left(\mathrm{t}\right)=\mathrm{2t}+\mathrm{5}\:\mathrm{then}\:,\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{at}\:\mathrm{4s}\:\mathrm{is}............ \\ $$

Commented by RasheedSindhi last updated on 21/May/17

velocity=x′(t)=2  Since velocity is constant  the acceleration is  0  (Also  at  4s)

$$\mathrm{velocity}=\mathrm{x}'\left(\mathrm{t}\right)=\mathrm{2} \\ $$$$\mathrm{Since}\:\mathrm{velocity}\:\mathrm{is}\:\mathrm{constant} \\ $$$$\mathrm{the}\:\mathrm{acceleration}\:\mathrm{is}\:\:\mathrm{0} \\ $$$$\left(\mathrm{Also}\:\:\mathrm{at}\:\:\mathrm{4s}\right) \\ $$

Commented by chintan vasoya last updated on 21/May/17

thanks

$$\mathrm{thanks} \\ $$

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