PYQ NEET- Motion In A Straight Line Kinematics L-6

Question: If the velocity of a particle is $v=A t+B t^2$, where $\mathrm{A}$ and $\mathrm{B}$ are constants, then the distance travelled by it between $1 \mathrm{~s}$ and $2 \mathrm{~s}$ is#

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

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

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

D) $3 A+7 B$

Answer: $\frac{3}{2} A+\frac{7}{3} B$#

Sol:#

Given, $v=A t+B t^2$ $$ \begin{aligned} & \frac{d x}{d t}=A t+B t^2 \ & \int d x=\int\left(A t+B t^2\right) d t \ & \mathrm{x}=\frac{A t^2}{2}+\frac{B t^3}{3}+C \end{aligned} $$

At $t=1$, particle is at $$ \mathrm{x}(\mathrm{t}=1)=\frac{A}{2}+\frac{B}{3}+C $$

At $t=2$, particle is at $$ \mathrm{x}(\mathrm{t}=2)=2 A+\frac{8 B}{3}+C $$ $\therefore$ distance travelled by the particle between $1 \mathrm{~s}$ and $2 \mathrm{~s}$ is, $$ \begin{aligned} & =\mathrm{x}(\mathrm{t}=2)-\mathrm{x}(\mathrm{t}=1) \ & =\left(2 A+\frac{8 B}{3}+C\right)-\left(\frac{A}{2}+\frac{B}{3}+C\right) \ & =\frac{3}{2} A+\frac{7}{3} B \end{aligned} $$