If the velocity of a particle is v = A t + B t 2 , where A and B are constants, then the distance travelled by it between 1 s and 2 s is: [2016]

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Solution

Q.
If the velocity of a particle is $v = A t + B t^{2},$ where A and B are constants, then the distance travelled by it between 1 s and 2 s is: [2016]

1 $\frac{3}{2} A + \frac{7}{3} B$

2 $\frac{A}{2} + \frac{B}{3}$

3 $\frac{3}{2} A + 4 B$

4 $3 A + 7 B$

Ans.
(1)

Velocity of the particle is $v = A t + B t^{2}$

$\frac{d s}{d t} = A t + B t^{2}; \int d s = \int (A t + B t^{2}) d t ∴ s = \frac{A t^{2}}{2} + B \frac{t^{3}}{3} + C$

$s (t = 1 s) = \frac{A}{2} + \frac{B}{3} + C; s (t = 2 s) = 2 A + \frac{8}{3} B + C$

$Required distance = s (t = 2 s) - s (t = 1 s)$

$= (2 A + \frac{8}{3} B + C) - (\frac{A}{2} + \frac{B}{3} + C) = \frac{3}{2} A + \frac{7}{3} B$

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