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 $t = 2\,s$ to $t = 6\,s$ is

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

Answer 1

Answer: (B)

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

$\therefore $ Total distance travelled, $S=S_1+S_2$.
During the time interval $0$ to $5 \,s$,
the acceleration of particle is equal to the slope of line $OA$
i.e. $a=\frac{12}{5}=2.4\,m\,s^{-2}$
Velocity at the end of $2 \,s$ will be,
$v = 0 + 2.4 \times 2 = 4.8 \,m \,s^{-1}$
Taking motion of particle for time interval $2\, s$ to $5 \,s$,
Here, $u = 4.8 \,m\, s^{-1}$, $a = 2.4 \,m \,s^{-2}$,
$S = S_1$, $t = 5 - 2 = 3\,s$
Then, $S_{1}=4.8 \times 3+\frac{1}{2}\times2.4\times3^{2}=25.2\,m$
Acceleration of the particle during the motion $t = 5 \,s$ to $t = 10\, s$ is
$a =$ Slope of line $AB=-\frac{12}{5}=-2.4\,m\,s^{-2}$
Taking motion of the particle for the time $1 \,s$ (i.e $5 \,s$ to $6\, s$),
Here, $u = 12\, m \,s^{-1}$, $a = - 2.4 \,m \,s^{-2}$, $t = 1 \,s$, $S = S_2$
$\therefore S_{2}=12\times1+\frac{1}{2}\left(-2.4\right)\times1^{2}=10.8\,m$
$\therefore S=S_{1}+S_{2}=25.2+10.8=36\,m$