Question

The v-t graph for a particle moving along x-axis is shown in the figure, find

(a) Average velocity in the interval 0 to 8 s.
(b) Average speed in the interval 0 to 8 s.
(c) Average acceleration in the interval 0 to 4 s.
(d) Instantaneous acceleration at t = 2.5 s.

Answer 1

(a) Displacement in the interval 0 to 8 s is equal to the area under the graph, taking the area below time axis as negative.
`rArr" Displacement"=(-(1)/(2)xx2xx8)+((1)/(2)xx6xx8)`
= 16 m
`"Average velocity"=("displacement")/("time interval")=(16)/(8)=2m//s`

(b) Distance in the interval of 0 8 s is equal to the area under the graph taking the both areas (above and below time axis) positive.
`"Distance"=((1)/(2)xx2xx8)+((1)/(2)xx6xx8)`
= 32 m
`"Average speed "=("distance")/("time interval")`
`=(32)/(8)=4m//s`
(c) `"Average acceleration"=("change in velocity")/("time interval")`
change in velocity in the interval 0 to 4 s
`=" (Velocity at t = 4 s)"-"(Velocity at t = 0 )"=8-(-8)=16m//s` ltbr. `rArr" Average acceleration"=(16)/(4)=4m//s^(2)`

(d)

Instantaneous acceleration at t = 3 s is the slope of graph at point
`P=tan theta=(8)/(2)=4m//s^(2)`