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Question Number 25103 by Tinkutara last updated on 03/Dec/17

Answered by prakash jain last updated on 05/Dec/17

$${\int }_0^5\mid 2t-3\mid dt$$$$={\int }_0^{3/2}(3-2t)dt+{\int }_{3/2}^5(2t-3)dt$$$$={[3t-t^2]}_0^{3/2}+{[t^2-3t]}_{3/2}^5$$$$=[\frac{9}{2}-\frac{9}{4}]+[25-15-\frac{9}{4}+\frac{9}{2}]$$$$=\frac{9}{4}+10+\frac{9}{4}$$$$=14.5\mathrm{m}$$

Commented by Tinkutara last updated on 05/Dec/17

$$Butitwillgivedistance,weneed$$$$displacementsowhynotjustfinal-$$$$initial?$$

Commented by prakash jain last updated on 05/Dec/17

$$\mathrm{Velocity}\mathrm{is}\mathrm{always}+\mathrm{ve}(given\mid 2t-3\mid )\mathrm{so}$$$$\mathrm{distance}=\mathrm{displacement}$$

Commented by ajfour last updated on 05/Dec/17

$$velocityfunctionshouldbe$$$$given.$$

Commented by prakash jain last updated on 05/Dec/17

$$\mathrm{it}\mathrm{is}\mathrm{moving}\mathrm{in}\mathrm{straight}\mathrm{line}\mathrm{and}$$$$\mathrm{velocity}\mathrm{is}\mathrm{always}+\mathrm{ve}\mathrm{as}\mathrm{v}=\mid 2t-3\mid .$$$$\mathrm{So}\mathrm{velocity}\mathrm{function}\mathrm{is}\mathrm{given}.$$

Commented by Tinkutara last updated on 06/Dec/17

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir}!$$

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