Answer :

(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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