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Home » If the velocity of a particle is , where and are constants, then the distance travelled by it between and is: A B C D
If the velocity of a particle is \mathrm{v}=\mathrm{At}+\mathrm{Bt}^{2}, where \mathrm{A} and \mathrm{B} are constants, then the distance travelled by it between 1 \mathrm{~s} and 2 \mathrm{~s} is:
A \quad \frac{3}{2} \mathrm{~A}+4 \mathrm{~B}
B \quad 3 \mathrm{~A}+7 \mathrm{~B}
C \quad \frac{3}{2} \mathrm{~A}+\frac{7}{3} \mathrm{~B}
D \quad \frac{\mathrm{A}}{2}+\frac{\mathrm{B}}{3}
Physics
If the velocity of a particle is \mathrm{v}=\mathrm{At}+\mathrm{Bt}^{2}, where \mathrm{A} and \mathrm{B} are constants, then the distance travelled by it between 1 \mathrm{~s} and 2 \mathrm{~s} is:
A \quad \frac{3}{2} \mathrm{~A}+4 \mathrm{~B}
B \quad 3 \mathrm{~A}+7 \mathrm{~B}
C \quad \frac{3}{2} \mathrm{~A}+\frac{7}{3} \mathrm{~B}
D \quad \frac{\mathrm{A}}{2}+\frac{\mathrm{B}}{3}

Correct Option C

Solution:
Given in question,

\mathrm{V}=\mathrm{At}+\mathrm{Bt}^{2}
We know that v = dx/dt
Thus,

\mathrm{x}= \int \mathrm{At}+\mathrm{Bt}^{2} \mathrm{dt}

\left[\frac{A t^{2}}{2}+\frac{B t^{3}}{3}\right]_{1}^{2}

=\mathrm{A}\left[\frac{4-1}{2}\right]+\mathrm{B}\left[\frac{8-3}{3}\right]

=\frac{3}{2} \mathrm{~A}+\frac{7}{3} \mathrm{~B}

i

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