# 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)?

# Answer:

(a) Distance travelled by the particle = Area
under the given graph

Average speed =

(b) Let s

_{1}and s_{2}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 = s

_{1}+ s_{2}… (i)
For distance s

_{1}:
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

v = u + at

Where,

v = Final velocity of the particle

12 = 0 + a′ × 5

12 = 0 + a′ × 5

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

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

Distance travelled by
the particle in 1s (i.e., between t = 5 s and t = 6 s)

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