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Motion in Straight Line

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12
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)?

a)

Distance travelled by the particle = Area under the given graph.

= (1/2) × (10 - 0) × (12 - 0)

= 60 m

b)

Let s1 and s2 be the distances covered by the particle between time = 2 s to 5 s and t = 5s to 6 s respectively.

Total distance (s) covered by the particle in time t = 2 s to 6 s is,

s = s1 + s2                     … (i)
For distance s1

Let u1 be the velocity of the particle after 2 s, and
a1 be the acceleration of the particle in t = 0 to t = 5s.

The particle undergoes uniform acceleration in the first 5 seconds.

Therefore, using first equation of motion v = u+at, we have

12 = 0 + a1 × 5

i.e.,  a1 = 12/5  = 2.4 m/s2

Again, using the first equation of motion, we have

v 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 = u1 t + (1/2) a1 t

= 4.8 × 3 + (1/2) × 2.4 × (3)

= 25.2 m                                   ...(ii)

For distance s2

Let a2 be the acceleration of the particle between time t = 5 s and t = 10 s.

Using first equation of motion,

v = u + at

0 = 12 + a2 × 5

a2 = -12 / 5 = - 2.4 ms-2

Therefore,

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

s2u2 t + (1/2)a2 t2

= 12 × 1 + (1/2) (-2.4) × (1)

= 12 - 1.2 = 10.8 m                 ...(iii)

Now, from equations (i), (ii) and (iii), we have

s = 25.2 + 10.8 = 36 m

∴ Average speed = 36 / 4 = 9 m/s

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