The speed-time graph of a particle moving along a fixed direction is shown in the Fig. 2 (CF) . 15. The distance traversed by the particle between ` t= 2 s` ` 6 s` is .
.

Options

A. ` 26 m`
B. ` 36 m`
C. ` 46 m`
D. `56 m`

Answer

B

Explanation

Let ` s_1` be the distace travelled by particle in time ` 2` ro `5` s and ` s-2` be the distance travelled by particle in time ` 5` to `6` s.
:. Total distance travelled, ` s= s_1 + s_2`.
. ,brgt During the tiem interval ` 0` to ` 5 s`, the accelration of particle is equal t the slope of ` OA i.e.
`a = (12) / 5 = 2.4 ms^(-2)`.
Velocity at the end of ` 2 s` will be ,
` v= 0 + 2.4 xx 2 = 4.8 ms^(-1)`
Taking motion of particle for time interval ` 2 s`. to
The , ` s_1 = 4 . 8 xx 3 + 1/2 2.4 xx 3^2 = 25 .2m`
Acceleration of the particle during the motion
` t= 5 s to ` t= 10 s` is
` a= slope of line ` AB =- (12)/5 =- 2. 4 ms^(-2)`
Taking motion of the particle for the time ` 1 sec` (i.e. 5 s to 6 s), we have
` u= 12 ms^(-1), a =-2.4 ms^(-2), t = 1 , s=s_2`
` :. s_2 = 12 xx 1 + 1/2 (-2 .4 ) xx 1^@ = 10.8 m`
:.` S= s-1 _ s_2 = 25. + 10 .8= 36 m`.