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 =0 s` to `10s. (b) t = 2 s ` to 6s.

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

Explanation

The distance travelled = Area under the ` v to t `graph.
For the first case,

Area = Area of `Delta OAB= 1/2 xx 10 xx 2 = 60 m`
Constant acceleratioon during
time 0 s to 5 s = slope of the graph OA
`therefore a = 12/5 = 2.4 ms ^(-2)`
Now velocity of a particle after t = 2 s is
`v _((2)) = v _(0) + at = 0 + 2.4 xx 2 = 4.8 ms ^(-1)`
For the distance travels between time t= 2 s to t = 5 s,
`d _(1) = v _((2)) xx (5-2) + 1/2 a (5-2) ^(2)`
`= (4.8 xx 3) + 1/2 xx 2.4 xx (3) ^(2) = 14.4 + 1.2 xx 9 = 14.4 +10.8 `
`therefore d _(1) = 25.2m " "...(1)`
Now, during motion on AB path, constant retardation a.= slope of the line AB
`therefore a. = (12)/(-5) =-2.4 ms ^(-2)`
Now time interval `t.=6 sec - 5 sec = 1 sec`
`d _(2) = v _(0) t + 1/2 a.t. ` when `v _(0) = 12 ms ^(-1)`
`d _(2) = (12 xx 1 ) + 1/2 (-2.4) xx 1`
`= 12 - 1.2 = 10.8 m`
If total distance travels by a particle during `t = 2` s to t = 6 s
`therefore d = d _(1) +d _(2)`
`= 25.2 +10.8`
`therefore d = 36.0 m`