# The speed-time graph of a particle moving along a fixed direction is

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

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 $=\frac{36}{4}=9 \mathrm{~m} / \mathrm{s}$