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)?

a)
Distance travelled by the particle = Area under the given graph.
                                           = (1/2) × (10 - 0) × (12 - 0)
                                           = 60 m


b) 
Let s₁ and s₂ be the distances covered by the particle between time t = 2 s to 5 s and t = 5s to 6 s respectively.
Total distance (s) covered by the particle in time t = 2 s to 6 s is, 
                           s = s₁ + s₂                     … (i) 
For distance s₁ , 
Let u₁ be the velocity of the particle after 2 s, and 
a₁ be the acceleration of the particle in t = 0 to t = 5s. 
The particle undergoes uniform acceleration in the first 5 seconds. 
Therefore, using first equation of motion v = u+at, we have
    12 = 0 + a₁ × 5
i.e.,  a₁ = 12/5  = 2.4 m/s²
Again, using the 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
s₁ = u₁ t + (1/2) a₁ t² 

   = 4.8 × 3 + (1/2) × 2.4 × (3)² 
 
   = 25.2 m                                   ...(ii) 

For distance s₂: 
Let a₂ be the acceleration of the particle between time t = 5 s and t = 10 s.
Using first equation of motion, 
v = u + at
0 = 12 + a₂ × 5
a₂ = -12 / 5 = - 2.4 ms^-2
Therefore, 
Distance travelled by the particle in 1s (i.e., between t = 5 s and t = 6 s),
s₂ = u₂ t + (1/2)a₂ t²
   = 12 × 1 + (1/2) (-2.4) × (1)² 

   = 12 - 1.2 = 10.8 m                 ...(iii) 
Now, from equations (i), (ii) and (iii), we have
s = 25.2 + 10.8 = 36 m
∴ Average speed = 36 / 4 = 9 m/s