Question

The speed-time graph of a particle moving along a fixed direction is shown in the figure. The distance traversed by the particle between $t=2s$ to $t=6s$ is

• A. $26m$
• B. $36m$
• C. $46m$
• D. $56m$

Answer 1

Answer: (B)

Let $S_1$ be the distance travelled by particle in time $2s$ to $5s$ and $S_2$ be the distance travelled by particle in time $5$ to $6s$.

$\therefore$ Total distance travelled, $S=S_1+S_2$.
During the time interval $0$ to $5s$,
the acceleration of particle is equal to the slope of line $OA$
i.e. $a=\frac{12}{5}=2.4m{s}^{-2}$
Velocity at the end of $2s$ will be,
$v=0+2.4\times 2=4.8m{s}^{-1}$
Taking motion of particle for time interval $2s$ to $5s$,
Here, $u=4.8m{s}^{-1}$, $a=2.4m{s}^{-2}$,
$S=S_1$, $t=5-2=3s$
Then, $S_1=4.8\times 3+\frac{1}{2}\times 2.4\times 3^2=25.2m$
Acceleration of the particle during the motion $t=5s$ to $t=10s$ is
$a=$ Slope of line $AB=-\frac{12}{5}=-2.4m{s}^{-2}$
Taking motion of the particle for the time $1s$ (i.e $5s$ to $6s$),
Here, $u=12m{s}^{-1}$, $a=-2.4m{s}^{-2}$, $t=1s$, $S=S_2$
$\therefore S_2=12\times 1+\frac{1}{2}\left(-2.4\right)\times 1^2=10.8m$
$\therefore S=S_1+S_2=25.2+10.8=36m$