First slide

Instantaneous acceleration

Question

If the velocity of a particle is $v = At + {Bt}^{2}$ , where A and B are constants, then the distance travelled by it between Is and 2s is [NEET 2016]

Difficult

A

3A + 7B
B

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

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

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

Solution

Velocity of the particle is given as $v = At + {Bt}^{2}$
where, A and B are constants.
$\Rightarrow \frac{d x}{d t} = A t + B t^{2}$    $(∵ v = \frac{d x}{d t})$
$\Rightarrow d x = (A t + B t^{2}) d t$
Integrating both sides, we get
$\int_{x_{1}}^{x_{2}} d x = \int_{1}^{2} (A t + B t^{2}) d t$
$\Rightarrow Δ x = x_{2} - x_{1} = A \int_{1}^{2} t d t + B \int_{1}^{2} t^{2} d t$
   $= A {[\frac{t^{2}}{2}]}_{1}^{2} + B {[\frac{t^{3}}{3}]}_{1}^{2}$
$= \frac{A}{2} (2^{2} - 1^{2}) + \frac{B}{3} (2^{3} - 1^{3})$
$∴$ Distance travelled by the particle between Is and 2s is
   $Δ x = \frac{A}{2} \times (3) + \frac{B}{3} (7) = \frac{3 A}{2} + \frac{7 B}{3}$

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