(a) Distance travelled by the particle is equal to the area under s-t graph.

∴ Distance covered in time interval t=0 to t=10 is,

D = 0.5×10×12 m

= 60 m

Average speed = m/s

= = 6 m/s

(b) Since, the particle is undergoing increase in speed in interval t=2 s to t=5 s and decrease in speed in interval t=5 s to t= 6 s. So, we have to deal this interval separately.

Let distance traveled in between 2 s and 5 s be d₁ and distance travelled in between 5 s and 6 s be d₂. Then,

Total distance covered in t=2 s and t=6 s be,

D = d₁ + d₂

For distance d₁:

For time interval t=0 s to t=5 s,

From 1^st equation of motion,

v = u + at

Where,

v = Final velocity = here 12 m/s

u = Initial velocity = here 0 m/s

a = Acceleration/Deceleration = Let a’

t = Time = 5 s

⇒12 = 0 + 5a’

∴ a’ = 2.4 m/s²

Again, from first equation of motion and t = 2 s

v’ = 0 + 2×2.4 = 4.8 m/s

Distance travelled by the particle in t=2 s to t=5s .i.e.,3s

From second equation of motion,

s = ut + 0.5at²

Where,

u = Initial velocity = v’ m/s = 4.8 m/s

a = Acceleration/Deceleration = 2.4 m/s²

s = Distance covered = d₁

t = Time = 3 s

⇒ d₁ = (4.8×3) + (0.5×2.4×3²) = 25.2 m

For distance d₂:

For time interval t=5 s to t=10 s,

Final velocity = velocity at 10 s = 0 m/s

Initial velocity = velocity at 5 s = 12 m/s

From 1^st equation of motion,

v = u + at

Where,

v = Final velocity = here 0 m/s

u = Initial velocity = here 12 m/s

a = Acceleration/Deceleration = Let a’

t = Time = 5 s

⇒ 0 = 12 + 5a’

∴ a’ = -2.4 m/s²

Distance travelled by the particle in t=5 s to t=6 s..i.e.,1s

From second equation of motion,

s = ut + 0.5at²

Where,

u = Initial velocity = v’ m/s = 12 m/s

a = Acceleration/Deceleration = -2.4 m/s²

s = Distance covered = d₁

t = Time = 1 s

⇒ d₂ = (12×1) + (0.5×-2.4×1²) = 10.8 m

∴ Total distance covered in t=2 s and t=6 s be,

D = d₁ + d₂ = 25.2 + 10.8 = 36 m

And, Average speed = m/s

= = 9 m/s