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(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 1st 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 1st 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

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