Question -
Answer -
(a) Distance traversedby the particle between t = 0 s and t = 10 s
= area of the triangle= (1/2) x base x height
= (1/2) x 10 x x12 =60 m
Average speed of theparticle is 60 m/ 10 s = 6 m/s
(b) The distancetravelled by the particle between t = 2 s and t = 6 s
= Let S1 bethe distance travelled by the particle in time 2 to 5 s and S2 bethe distance travelled by the particle in time 5 to 6 s.
For the motion from 0sec to 5 sec
Now, u = 0 , t = 5 , v= 12 m/s
From the equation v =u + at we get
a = (v – u)/t = 12/ 5= 2.4 m/s2
Distance covered from2 to 5 s, S1 = distance covered in 5 sec – distance covered in 2 sec
= (1/2) a (5)2 –(1/2) a (2)2 = (1/2) x 2.4 x (25 – 4) = 1.2 x 21 = 25.2 m
For the motion from 5sec to 10 sec , u = 12 m/s and a = -2.4 m/s2
and t = 5 sec to t = 6sec means n = 1 for this motion
Distance covered inthe 6 the sec is S2 = u + (1/2) a (2n – 1)
= 12 – (2.4/2) (2 x 1– 1) = 10.8 m
Therefore, the totaldistance covered from t = 2 s to 6 s = S1 + S2
= 25.2 + 10.8 = 36 m