The speed-time graph of a particle moving along a fixed direction is shown in the Fig. 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)?

Explanation

(a) Distance travelled by the particle = Area under the given graph
`=1/2xx(10-0)xx(12-0)=60` m
Average speed =`("Distance")/("Time")=60/10` =6 m/s
(b) 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 = s_1 + s_2 … (i)`
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′ × 5
`d'=12/5=2.4 m//s^2`
Again, from first equation of motion, we have
v = u + at
= 0 + 2.4 × 2 = 4.8 m/s
Distance travelled by the particle between time 2 s and 5 s i.e., in 3 s
`s_1=u't+1/2a't^2`
`=4.8xx3+1/2xx2.4xx(3)^2`
=25.2 m ....(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″ × 5
`a''=(-12)/5`
`=-2.4 m//s^2`
Distance travelled by the particle in 1s (i.e., between t = 5 s and t = 6 s)
`s_2=u''t+1/2at^2`
`=12xxa+1/2(-2.4)xx(1)^2`
=12-1.2=10.8 m .....(iii)
From equation (i),(ii) and (iii) , we get
s = 25.2 + 10.8 = 36 m
`therefore` Average speed `=36/4=9` m/s