The speed -time graph of a particle moving along a fixed direction id shown ifn Fig. 2 (b) . 31. Fid (i) distance travelled by the particle between ` 0 sec` to ` 10 sec` (ii) average speed between thid interval (iii) the time when the speed was minimum (iv) the time when the speed was maximum.
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Explanation

(a) Distanc travelled by the particle between ` 0` to ` 10` s will be =area of ` Delta OAB`, whose base of ` 10 s` and height is `12 ms^(-1)`
` =1/2 xx 10 xx 12 = 60 m`
Average speed ` = (60)/(10) = 6 ms^(-1)`
(b) Let `S_1` and `S_2` be the distances coverd by the particle in the time in terval ` t_1 =2 s to ` 5 s` and ` t_2 = 5 s` to 6 s, then total distance covered in time inerval `t=2 to ` 6 s` will be ` S= S_1` ....(i)
To finds ` S_1`, let us consider ` u_1` is the velocity of particle agter ` 2 secpmds` and ` a_1` is the accelration of the particle during the time interval zero to ` 5 seocond` . ltbRgt The ` u= 0, = 12 m//s, a= a_1 ` and ` t= 5 s`. ltbRgt We have, `a_1 = (v-u)/t = ( 12-0)/5 = (120/5 = 2.4 ms^(-20` :. ` u-1 = u + a_1 t= 0 + 2.4 xx 2 = 4.8 ms^(-1)`
Thus for tje distanc etravelled by particle in `3 secons` (i.e. time inrerval ` 2 s` to 5 s) we have ltbRgt ` u_1 = 4.8 ms^(-1), t-1 =3 s, a_1 = 2.4 ms^(-20, S_1 = ?`
As ` S_1 = u-1 t_1 = 1/2 a_1 r_1^2` :. ` S-1 = 4.8 xx 3 + 1/2 xx 2. 4 xx 3^2 = 25 .2 m`
To find ` S_2`. Let ` a-2` be the the accelration of the particle during the motion, ` t= 5 to ` t= 10 s`. We have
` a_2 = ( 0-12)/( 10- 5) =- 2. 4 ms^(2)`
Taking motion of the particle in time interval ` t= 5 s tp ` t=6 s`, we have ltbRgt ` u_2 = 12 ms^(-1), a-2 =- 2.4 ms^(-2) , t_2 = 1 s, S_2 = ?`
As ` S-2 = u-2 t_2 + 1/2 a_2 t_2^2` :. ` S_2 = 12 xx 1 = 1/2 (- 2. 4) 1^2 = 10.8 m`
:. Total distance travelled, `S= 25. + 10 .8 = 36 m`
Acerage speed = 936)/( 6- 2) = 9360/4 = 9 ms^(-1)`.