The speed-time graph of a particle moving along a fixed direction is shown in the figure. The distan

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

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(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 . ∴ 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 $O A$\ $ 	ext { i.e. } \quad a=\frac{12}{5}=2.4 \mathrm{~m} \mathrm{~s}^{-2} $\ Velocity at the end of 2 s will be, $v=0+2.4 	imes 2=4.8 \mathrm{~m} \mathrm{~s}^{-1}$ Taking motion of particle for time interval 2 s to 5 s , Here, $u=4.8 \mathrm{~m} \mathrm{~s}^{-1}, a=2.4 \mathrm{~m} \mathrm{~s}^{-2}, S=S_1, t=5-2=3 \mathrm{~s}$ Then, $S_1=4.8 	imes 3+\frac{1}{2} 	imes 2.4 	imes 3^2=25.2 \mathrm{~m}$ Acceleration of the particle during the motion $t=5 \mathrm{~s}$ to $t=10 \mathrm{~s}$ is\ $ a=	ext { Slope of line } A B=-\frac{12}{5}=-2.4 \mathrm{~m} \mathrm{~s}^{-2} $ \Taking motion of the particle for the time 1 s (i.e 5 s to 6 s ), Here, $u=12 \mathrm{~m} \mathrm{~s}^{-1}, a=-2.4 \mathrm{~m} \mathrm{~s}^{-2}, t=1 \mathrm{~s}, S=S_2$ $ \ \begin{aligned} & 	herefore S_2=12 	imes 1+\frac{1}{2}(-2.4) 	imes 1^2=10.8 \mathrm{~m} \ & 	herefore S=S_1+S_2=25.2+10.8=36 \mathrm{~m} \end{aligned} $