First slide

Displacement in nth sec

Question

If the velocity of a particle is $v = A t + B t^{2},$ where $A a n d B$ are constants, then the distance travelled by it between $1 s a n d 2 s$ is

Moderate

A

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

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

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

$3 A + 7 B$

Solution

$Velocity of the particle is v = A t + B t^{2}$
$\frac{d s}{d t} = A t + B t^{2}, \int d s = \int (A t + B t^{2}) d t$
$\begin{array}{l} ∴ s = \frac{A t^{2}}{2} + B \frac{t^{3}}{3} + C \\ s (t = 1 s) = \frac{A}{2} + \frac{B}{3} + C \\ s (t = 2 s) = 2 A + \frac{8}{3} B + C \end{array}$
$Required distance = s (t = 2 s) - s (t = 1 s)$
$= (2 A + \frac{8}{3} B + C) - (\frac{A}{2} + \frac{B}{3} + C) = \frac{3}{2} A + \frac{7}{3} B$

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