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If the velocity of a particle is \mathrm{V=A t^{2}+B t^{3}} where \mathrm{A} and \mathrm{B} are constants, then the distance travelled by it between 1s and 25 is-

Option: 1

\mathrm{\frac{7 A}{3}+\frac{15 B}{4}}


Option: 2

\mathrm{\frac{7 A}{2}+\frac{15 B}{3}}


Option: 3

\mathrm{\frac{7 A}{4}+\frac{15 B}{4}}


Option: 4

\mathrm{\frac{7 A}{4}+\frac{15 B}{4}}


Answers (1)

best_answer

Velocity equation is given by-

\mathrm{V=A t^{2}+B t^{3}}\quad (1)

To find the distance of the particle we have to integrate it
\mathrm{\text { Distance }(s) =\int_{1 S}^{2 S}\left(A t^{2}+B t^{3}\right) d t}
\mathrm{s =\left[\frac{A t^{3}}{3}+\frac{B t^{4}}{4}\right]_{ 1 S}^{2 S} }
\mathrm{s =\left[\frac{A}{3} \times 8+\frac{B}{4} \cdot 16\right]-\left[\frac{A}{3}+\frac{B}{4}\right]}
\mathrm{s =\frac{8 A}{3}+\frac{16 B}{4}-\frac{A}{3}-\frac{B}{4}}

\mathrm{s =\frac{8 A-A}{3}+\frac{16 B-B}{4}= \frac{7A}{3}+\frac{15B}{4}}
\mathrm{S= \frac{7 A}{3}+\frac{15 B}{4} }

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Rakesh

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