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.
(Fig. 3.28)
What is the average speed of the particle over the intervals in (a) and (b)?
(a) Distance travelled by the particle = Area under the given graph
= ½ 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 m/s2
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
= 4.8x3 + 1/2 x 2.4 x (3)2
= 25.2m ....(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 m/s2
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 a + ½ (-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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