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If the velocity of a particle is $v = At + Bt^2$, where A and B are constants, then the distance travelled by it between 1 s and 2 s is
A
3 A + 7 B
B
$\frac{3}{2}A + \frac{7}{3}B$
C
$\frac{A}{2} + \frac{B}{3}$
D
$\frac{3}{2}A + 4B$
Explanation
Integrate v(t) between the limits.
Detailed Solution
$v = \frac{dx}{dt} \Rightarrow x = \int_1^2(At + Bt^2)dt = \left[\frac{At^2}{2} + \frac{Bt^3}{3}\right]_1^2$
$x = \frac{4A}{2} + \frac{8B}{3} - \frac{A}{2} - \frac{B}{3}$
$x = \frac{3A}{2} + \frac{7B}{3}$
$x = \frac{4A}{2} + \frac{8B}{3} - \frac{A}{2} - \frac{B}{3}$
$x = \frac{3A}{2} + \frac{7B}{3}$