If the velocity of a particle is v = At + Bt^2, where A and B are constants, then the distance travelled by it b/w 1 s and 2 s is
1) A/2 + B/3
2) 3/2×A + 4B
3) 3A + 7B
4) 3/2×A + 7/3B
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403
the answer is option D
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Answered by
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velocity of a particle is given by
![\qquad v=At+Bt^2\\\\\\\frac{dx}{dt}=At+Bt^2\\\\\\dx=(At+Bt^2).dt\\\\\\\int\limits^x_0{dx}=\int\limits^2_1{(At+Bt^2)}\,dt\\\\\\x=A\left[\frac{t^2}{2}\right]^2_1+B\left[\frac{t^3}{3}\right]^2_1\\\\\\x=A\left(\frac{2^2}{2}-\frac{1^2}{2}\right)+B\left(\frac{2^3}{3}-\frac{1^3}{3}\right)\\\\\\x=\frac{3}{2}A+\frac{7}{3}B](https://tex.z-dn.net/?f=%5Cqquad+v%3DAt%2BBt%5E2%5C%5C%5C%5C%5C%5C%5Cfrac%7Bdx%7D%7Bdt%7D%3DAt%2BBt%5E2%5C%5C%5C%5C%5C%5Cdx%3D%28At%2BBt%5E2%29.dt%5C%5C%5C%5C%5C%5C%5Cint%5Climits%5Ex_0%7Bdx%7D%3D%5Cint%5Climits%5E2_1%7B%28At%2BBt%5E2%29%7D%5C%2Cdt%5C%5C%5C%5C%5C%5Cx%3DA%5Cleft%5B%5Cfrac%7Bt%5E2%7D%7B2%7D%5Cright%5D%5E2_1%2BB%5Cleft%5B%5Cfrac%7Bt%5E3%7D%7B3%7D%5Cright%5D%5E2_1%5C%5C%5C%5C%5C%5Cx%3DA%5Cleft%28%5Cfrac%7B2%5E2%7D%7B2%7D-%5Cfrac%7B1%5E2%7D%7B2%7D%5Cright%29%2BB%5Cleft%28%5Cfrac%7B2%5E3%7D%7B3%7D-%5Cfrac%7B1%5E3%7D%7B3%7D%5Cright%29%5C%5C%5C%5C%5C%5Cx%3D%5Cfrac%7B3%7D%7B2%7DA%2B%5Cfrac%7B7%7D%7B3%7DB "\qquad v=At+Bt^2\\\\\\\frac{dx}{dt}=At+Bt^2\\\\\\dx=(At+Bt^2).dt\\\\\\\int\limits^x_0{dx}=\int\limits^2_1{(At+Bt^2)}\,dt\\\\\\x=A\left[\frac{t^2}{2}\right]^2_1+B\left[\frac{t^3}{3}\right]^2_1\\\\\\x=A\left(\frac{2^2}{2}-\frac{1^2}{2}\right)+B\left(\frac{2^3}{3}-\frac{1^3}{3}\right)\\\\\\x=\frac{3}{2}A+\frac{7}{3}B")
hence, option (4) is correct.
hence, option (4) is correct.
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