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

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

Explanation

(a) Distance covered by the particle in 10s
=Area enclosed by the graph
`=(1)/(2)` Base `xx` Height
`=(1)/(2)xx10xx12=60m`
Average speed between t=0 and t=10s
`=("Total path length")/("Time")=(60)/(10)=6m//s`
(b) So, distance travelled from t=2 s to t=6s= distance travelled from t=2 s to t=5s `(S_(1))+` distance travelled from t=5s to t=6 s `(S_(2))` Let `a_(1)`= acceleration of particle from t=0 to t=5sec.
From t=0to t=5s, final velocity, `v=12ms^(-1)`
u=0
`therefore v=u+a_(1)t`
Thus velocity after t=2s
v=u+at
`=0+2.4xx2=4.8ms^(-1)`
To find distance `S_(1)` travelled between t=2s and t=5 s we have
`Deltat=3s, a_(1)=2.4ms^(-2),v=4.8ms^(-1)`
`S_(1)=vDeltat+(1)/1(2)(2/4)xx(3)^(2)=25.2m`
To find `S_(2)`: (between t=5s to t=6s),
`a_(2)-(0-12)/(10-5)=-2.4ms^(-2)`
For t=5s to t=6s we have
`Deltat=1s, a_(2)=-2.4ms^(-2), v]=12ms^(-1)`
`therefore S_(2)=vt+(1)/(2)aDeltat^(2)rArr 12xx1+(1)/(2)(-2.4)xx1^(2)`
`S_(2)=10.8m`
Therefore, total distance travelled between t=2 s and t=6s is `S_(1)+S_(2)=25.2+10.8=36m`
Thus Average speed `=(36)/(6-2)=9ms^(-1)`