### Q.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) ?

**Answer:-**

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

= (1/2) × (10 – 0) × (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 × 3 + (1/2) × 2.4 × (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 × 1 + (1/2) (-2.4) × (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