The speed-time graph of a particle moving alo | vedclass

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Question

Let $S_{1}$ be the distance travelled by particle in
time $2$ to $5\, s$ and $S_{2}$ be the distance travelled
by particle in time $5$ to $6\, s.$

Vedclass · Accepted Answer

$36$

Vedclass · Answer

Let $S_{1}$ be the distance travelled by particle in
time $2$ 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 $OA$
i.e., $a=\frac{12}{5}=2.4 \mathrm{m} \mathrm{s}^{-2}$
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 $5 \mathrm{s}$
Here, $u=4.8 \mathrm{m} \mathrm{s}^{-1}, \mathrm{a}=2.4 \mathrm{m} \mathrm{s}^{-2}, \mathrm{S}=\mathrm{S}_{1}$
$t=5-2=3 s$
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
$t=5 s$ to $t=10 s$ is
$a=$ Slope of line $A B=-\frac{12}{5}=-2.4 \mathrm{ms}^{2}$
Taking motion of the particle for the time $1\, s$
$(\mathrm{ie}, 5 \mathrm{s} 	ext { to } 6 \mathrm{s}),$ Here $\mathrm{u}=12 \mathrm{m} \mathrm{s}^{-1}, \mathrm{a}=2.4 \mathrm{m} \mathrm{s}^{-2}$
$t=1 s, S=S_{2}$
$	herefore S_{2}=12 	imes 1+\frac{1}{2}(-2.4) 	imes 1^{2}=10.8 \mathrm{m}$
$	herefore \mathrm{S}=\mathrm{S}_{\mathrm{t}}+\mathrm{S}_{2}=25.2+10.8=36 \mathrm{m}$

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........$m$

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