Question.
The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) $t=0$ s to $10 \mathrm{~s}$, (b) $t=2 \mathrm{~s}$ to 6
The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) $t=0$ s to $10 \mathrm{~s}$, (b) $t=2 \mathrm{~s}$ to 6
solution:
(a) Distance travelled by the particle = Area under the given graph
$=\frac{1}{2} \times(10-0) \times(12-0)=60 \mathrm{~m}$
Average speed $=\frac{\text { Distance }}{\text { Time }}=\frac{60}{10}=6 \mathrm{~m} / \mathrm{s}$
(b) Let $s_{1}$ and $s_{2}$ be the distances covered by the particle between time
$t=2 \mathrm{~s}$ to $5 \mathrm{~s}$ and $t=5 \mathrm{~s}$ to $6 \mathrm{~s}$ respectively.
Total distance $(s)$ covered by the particle in time $t=2 \mathrm{~s}$ to $6 \mathrm{~s}$
$s=s_{1}+s_{2} \ldots$ (i)
For distance $s_{1}$ :
Let $u^{\prime}$ be the velocity of the particle after $2 \mathrm{~s}$ and $a^{\prime}$ be the acceleration of the particle in $t=0$ to $t=5 \mathrm{~s}$.
Since the particle undergoes uniform acceleration in the interval $t=0$ to $t=5 \mathrm{~s}$, from first equation of motion, acceleration can be obtained as:
$v=u+a t$
Where,
v = Final velocity of the particle
$12=0+a^{\prime} \times 5$
$a^{\prime}=\frac{12}{5}=2.4 \mathrm{~m} / \mathrm{s}^{2}$
Again, from first equation of motion, we have
$v=u+a t$
$=0+2.4 \times 2=4.8 \mathrm{~m} / \mathrm{s}$
Distance travelled by the particle between time 2 s and 5 s i.e., in 3 s
$s_{1}=u^{\prime} t+\frac{1}{2} a^{\prime} t^{2}$
$=4.8 \times 3+\frac{1}{2} \times 2.4 \times(3)^{2}$
$=25.2 \mathrm{~m}$ $\ldots$ (ii)
For distance $s_{2}$ :
Let $a^{\prime \prime}$ be the acceleration of the particle between time $t=5 \mathrm{~s}$ and $t=10 \mathrm{~s}$.
$s_{1}=u^{\prime} t+\frac{1}{2} a^{\prime} t^{2}$
$=4.8 \times 3+\frac{1}{2} \times 2.4 \times(3)^{2}$
$0=12+a^{\prime \prime} \times 5$
$a^{\prime \prime}=\frac{-12}{5}$
$=-2.4 \mathrm{~m} / \mathrm{s}^{2}$
Distance travelled by the particle in $1 \mathrm{~s}$ (i.e., between $t=5 \mathrm{~s}$ and $t=6 \mathrm{~s}$ )
$s_{2}=u^{\prime \prime} t+\frac{1}{2} a t^{2}$
$=12 \times a+\frac{1}{2}(-2.4) \times(1)^{2}$
$=12-1.2=10.8 \mathrm{~m}$ ... (iii)
From equations (i), (ii), and (iii), we get
$s=25.2+10.8=36 \mathrm{~m}$
$\therefore$ Average speed $=\frac{36}{4}=9 \mathrm{~m} / \mathrm{s}$
(a) Distance travelled by the particle = Area under the given graph
$=\frac{1}{2} \times(10-0) \times(12-0)=60 \mathrm{~m}$
Average speed $=\frac{\text { Distance }}{\text { Time }}=\frac{60}{10}=6 \mathrm{~m} / \mathrm{s}$
(b) Let $s_{1}$ and $s_{2}$ be the distances covered by the particle between time
$t=2 \mathrm{~s}$ to $5 \mathrm{~s}$ and $t=5 \mathrm{~s}$ to $6 \mathrm{~s}$ respectively.
Total distance $(s)$ covered by the particle in time $t=2 \mathrm{~s}$ to $6 \mathrm{~s}$
$s=s_{1}+s_{2} \ldots$ (i)
For distance $s_{1}$ :
Let $u^{\prime}$ be the velocity of the particle after $2 \mathrm{~s}$ and $a^{\prime}$ be the acceleration of the particle in $t=0$ to $t=5 \mathrm{~s}$.
Since the particle undergoes uniform acceleration in the interval $t=0$ to $t=5 \mathrm{~s}$, from first equation of motion, acceleration can be obtained as:
$v=u+a t$
Where,
v = Final velocity of the particle
$12=0+a^{\prime} \times 5$
$a^{\prime}=\frac{12}{5}=2.4 \mathrm{~m} / \mathrm{s}^{2}$
Again, from first equation of motion, we have
$v=u+a t$
$=0+2.4 \times 2=4.8 \mathrm{~m} / \mathrm{s}$
Distance travelled by the particle between time 2 s and 5 s i.e., in 3 s
$s_{1}=u^{\prime} t+\frac{1}{2} a^{\prime} t^{2}$
$=4.8 \times 3+\frac{1}{2} \times 2.4 \times(3)^{2}$
$=25.2 \mathrm{~m}$ $\ldots$ (ii)
For distance $s_{2}$ :
Let $a^{\prime \prime}$ be the acceleration of the particle between time $t=5 \mathrm{~s}$ and $t=10 \mathrm{~s}$.
$s_{1}=u^{\prime} t+\frac{1}{2} a^{\prime} t^{2}$
$=4.8 \times 3+\frac{1}{2} \times 2.4 \times(3)^{2}$
$0=12+a^{\prime \prime} \times 5$
$a^{\prime \prime}=\frac{-12}{5}$
$=-2.4 \mathrm{~m} / \mathrm{s}^{2}$
Distance travelled by the particle in $1 \mathrm{~s}$ (i.e., between $t=5 \mathrm{~s}$ and $t=6 \mathrm{~s}$ )
$s_{2}=u^{\prime \prime} t+\frac{1}{2} a t^{2}$
$=12 \times a+\frac{1}{2}(-2.4) \times(1)^{2}$
$=12-1.2=10.8 \mathrm{~m}$ ... (iii)
From equations (i), (ii), and (iii), we get
$s=25.2+10.8=36 \mathrm{~m}$
$\therefore$ Average speed $=\frac{36}{4}=9 \mathrm{~m} / \mathrm{s}$
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