## In This post you will get answer of Q.

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

Solution:

## 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, $S_{1}=4.8 times 3+frac{1}{2}times2.4times3^{2}=25.2,m$

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, $u = 12, m ,s^{-1}$, $a = – 2.4 ,m ,s^{-2}$, $t = 1 ,s$, $S = S_2$

$therefore S_{2}=12times1+frac{1}{2}left(-2.4right)times1^{2}=10.8,m$

$therefore S=S_{1}+S_{2}=25.2+10.8=36,m$