The speed-time graph of a particle moving along a fixed direction is shown in Fig. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2s to 6s

What is the average speed of the particle over the intervals in (a) and (b) ?

Explanation

a) Distance travelled by the particle between 0 to 10s will be
= Area of `DeltaOAB`, whose base is 10s and height is 12 m/s
`=1/2 times 10 times 12=60m`
Average speed `=60/10=6mS^(-1)`
b) Let `s_(1) and s_(2)` be the distances covered by the particle in the time interval `t_(1)=2s` to 5s and `t_(2)=5s" to "6s`, then total distance covered in time interval t = 2s to 6s will be `s=s_(1)+s_(2)" ..... (i) "`
To find `s_(1):` let us consider `u_(1)` is the velocity of particle after 2 seconds and `a_(1)` is the acceleration of the particle during the time interval zero to 5 seconds.
Then `u_(1)=0, v=12m//s, a=a_(1) and t=5s`
We have `a_(1)=(v-u)/t=(12-0)/5=12/5`
`" "=2.4m//s^(2)`
`therefore u_(1)=upsilon+a_(1)t=0+2.4 times 2=4.8m//s^(-1)`
Thus for the distance travelled by particle in 3 seconds (i.e, time interval 2s to 5s), we have
`u_(1)=4.8m//s, t_(1)=3s, a_(1)=2.4m//s^(2), s_(1)=?`
As `s_(1)=u_(1)t_(1)+1/2a_(1)t_(1)""^(2)`
`S_(1)=4.8+3 times 1/2 times 2.4 times 3^(2)=25.2m`
To find `s_(2):` let `a_(2)` be the acceleration of the particle during the motion
`t=5s" to "t=10s`
We have `a_(2)=(0-12)/(10-5)=-2.4m//s^(2)`
Taking motion of the partice in time interval
`" "t=5s" to "t=6s`
We have
`u_(1)=12m//s^(-1), a_(2)=-2.4m//s^(2), t_(2)=1s, s_(2)=?`
As `s_(2)=u_(2)t+1/2a_(2)t_(2)""^(2)`
`s_(2)=12 times 1+1/2(-2.4)1^(2)=10.8m`
`therefore` Total distance travelled s = 25.2 + 10.8 = 36 m
Average velocity `= (36)/(6-2)=36/4=9m//s`