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

Options

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

Answer

B

Explanation

Let `S_1` be the distance travelled by particle in time 2 to 5 s and `S_2` be the distance travelled by particle in time 5 to 6 s.
Time distance travelled,
`S = S_1 + S_2`.
During the time interval 9 to 5 s, the acceleration of particle is equal to the slope of line OA
i.e. `a = 12/5 = 2.4 m s^(-2)`
Velocity at the end of 2 s will be,
`v = 0 + 2.4 xx 2 = 4.8 m s^(-1)`
Taking motion of particle for time interval 2 s to 5 s,
Here, `u = 4.8 m s^(-1), a = 2.4 m s^(-2), S = S_1, t = 5 - 2 = 3 s`
Then, `S_1 = 4.8 xx 3 + 1/2 xx 2.4 xx 3^(2)` = 25.2 m
Acceleration of the particle during the motion t = 5 s to t = 10 s is
a = Slope of line `AB = - 12/5 = - 2.4 m s^(-2)`
Taking motion of the particle for the time 1 s (i.e 5 s to 6 s), Here, u = `12 m s^(-2), t = 1 s, S = S_2`
`therefore` `S_2 = 12 xx 1 + 1/2 (-2.4) xx 1^(2)` = 10.8 m
`therefore` S = `S_1 + S_2 = 25.2 + 10.8 = 36 m`