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 s, (b) t = 2 s to 6 s.

What is the average speed of the particle over the intervals in (a) and (b)?

Answer 1

Distance travelled by the particle = Area under the given graph
$=\frac{1}{2}\times (10-0)\times (12-0)=60mAveragespeed=\frac{Distance}{Time}=\frac{60}{10}=6m/s$
Let $s_1$ and $s_2$ be the distances covered by the particle between time
t = 2 s to 5 s and t = 5 s to 6 s respectively.
Total distance (s) covered by the particle in time t = 2 s to 6
$s=s_1+s_2\dots \left(i\right)$
For distance $s_1$:
Let u' be the velocity of the particle after 2 s and a' be the acceleration of the particle in t = 0 to t = 5 s.
Since the particle undergoes uniform acceleration in the interval t = 0 to t = 5 s, from first equation of motion, acceleration can be obtained as:
$v=u+at$

Where,
v = Final velocity of the particle
$12=0+a^{\prime }\times 5a^{\prime }=\frac{12}{5}=2.4m/s^2$
Again, from first equation of motion, we have
$v=u+at=0+2.4\times 2=4.8m/s$
Distance travelled by the particle between time 2 s and 5 s i.e., in 3 s
$s_1=u^{\prime }t+\frac{1}{2}{a}^{\prime }{t}^2=4.8\times 3+\frac{1}{2}\times 2.4\times (3{)}^2=25.2m\dots (ii)$
For distance $s_2$:
Let a" be the acceleration of the particle between time t = 5 s and t = 10 s.
From first equation of motion, v = u + at (where v = 0 as the particle finally comes to rest)
$0=12+a"\times 5a"=\frac{-12}{5}=-2.4m/s^2$
Distance travelled by the particle in 1s (i.e., between t = 5 s and t = 6 s)
$s_2=u"t+\frac{1}{2}{a}^{\prime \prime }{t}^2=12\times 1+\frac{1}{2}\times (-2.4)\times (1{)}^2=12-1.2=10.8\dots (iii)$
From equations (i), (ii), and (iii), we get
$s=25.2+10.8=36m\therefore Averagespeed=\frac{36}{4}=9m/s$