__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

*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*

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

s

_{1}= u' t + (1/2)a' t

^{2}

= 4.8 X 3 + (1/2) X 2.4 X (3)

^{2}

= 25.2 m ........(ii)

*For distance s*

_{2}

*:*

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)

s

_{2}= u" t + (1/2)

*a*″ t

^{2}

= 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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