Motion in a straight line NCERT Solutions Class 11 Physics - Solved Exercise Question 3.27

Question 3.27

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) ?


Solution:

(a) Distance travelled by the particle = Area under the given graph
= (1/2) X (10 - 0) X (12 - 0) = 60 m
Average speed = Distance / Time = 60 / 10 = 6 m/s

(b) Let s1 and s2 be the distances covered by the particle between time
t = 2 s to 5 s and t = 5 s to 6 s respectively.
Total distance (s) covered by the particle in time t = 2 s to 6 s
s = s1 + s2 … (i)

For distance s1:
Let u′ be the velocity of the particle after 2 s and a′ be the acceleration of the particle in t = 0 to t = 5 s.
Since the particle undergoes uniform acceleration in the interval t = 0 to t = 5 s, from first equation of motion, acceleration can be obtained as:
v = u + at
Where,
v = Final velocity of the particle

12 = 0 + a′ × 5
a′ = 12 / 5 = 2.4 ms-2
Again, from 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 = u' t + (1/2)a' t2
= 4.8 X 3 + (1/2) X 2.4 X (3)2
= 25.2 m      ........(ii)

For distance s2:
Let a″ be the acceleration of the particle between time t = 5 s and t = 10 s.
From first equation of motion,
v = u + at (where v = 0 as the particle finally comes to rest)
0 = 12 + a″ × 5
a″ = -12 / 5 = - 2.4 ms-2
Distance travelled by the particle in 1s (i.e., between t = 5 s and t = 6 s)
s2 = u" t + (1/2)a″ t2
= 12 X 1 + (1/2) (-2.4) X (1)2
= 12 - 1.2 = 10.8 m    .........(iii)

From equations (i), (ii), and (iii), we get
s = 25.2 + 10.8 = 36 m

∴ Average speed = 36 / 4 = 9 m/s

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