Question Number 25103 by Tinkutara last updated on 03/Dec/17
Answered by prakash jain last updated on 05/Dec/17
![∫_0 ^5 ∣2t−3∣dt =∫_0 ^(3/2) (3−2t)dt+∫_(3/2) ^5 (2t−3)dt =[3t−t^2 ]_0 ^(3/2) +[t^2 −3t]_(3/2) ^5 =[(9/2)−(9/4)]+[25−15−(9/4)+(9/2)] =(9/4)+10+(9/4) =14.5 m](Q25162.png)
$$\int_{\mathrm{0}} ^{\mathrm{5}} \mid\mathrm{2}{t}−\mathrm{3}\mid{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{3}/\mathrm{2}} \left(\mathrm{3}−\mathrm{2}{t}\right){dt}+\int_{\mathrm{3}/\mathrm{2}} ^{\mathrm{5}} \left(\mathrm{2}{t}−\mathrm{3}\right){dt} \\ $$$$=\left[\mathrm{3}{t}−{t}^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{3}/\mathrm{2}} +\left[{t}^{\mathrm{2}} −\mathrm{3}{t}\right]_{\mathrm{3}/\mathrm{2}} ^{\mathrm{5}} \\ $$$$=\left[\frac{\mathrm{9}}{\mathrm{2}}−\frac{\mathrm{9}}{\mathrm{4}}\right]+\left[\mathrm{25}−\mathrm{15}−\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{9}}{\mathrm{2}}\right] \\ $$$$=\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{10}+\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$=\mathrm{14}.\mathrm{5}\:\mathrm{m} \\ $$
Commented by Tinkutara last updated on 05/Dec/17
$${But}\:{it}\:{will}\:{give}\:{distance},\:{we}\:{need} \\ $$$${displacement}\:{so}\:{why}\:{not}\:{just}\:{final}− \\ $$$${initial}? \\ $$
Commented by prakash jain last updated on 05/Dec/17
$$\mathrm{Velocity}\:\mathrm{is}\:\mathrm{always}\:+\mathrm{ve}\:\left({given}\:\mid\mathrm{2}{t}−\mathrm{3}\mid\right)\mathrm{so} \\ $$$$\mathrm{distance}=\mathrm{displacement} \\ $$
Commented by ajfour last updated on 05/Dec/17
$${velocity}\:{function}\:{should}\:{be} \\ $$$${given}.\: \\ $$
Commented by prakash jain last updated on 05/Dec/17
$$\mathrm{it}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{and} \\ $$$$\mathrm{velocity}\:\mathrm{is}\:\mathrm{always}\:+\mathrm{ve}\:\mathrm{as}\:\mathrm{v}=\mid\mathrm{2}{t}−\mathrm{3}\mid. \\ $$$$\mathrm{So}\:\mathrm{velocity}\:\mathrm{function}\:\mathrm{is}\:\mathrm{given}. \\ $$
Commented by Tinkutara last updated on 06/Dec/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}! \\ $$