Question

Velocity-time graph of a particle moving in a straight line is shown in figure. At time `t= 0,` displacement of the particle from mean position is 10 m. Find

(a) acceleration of particle at `t = 1 s, 3 s and 9 s.`
(b) position of particle from mean position at `t = 10 s.`
(c) write down s-t equation for time interval (i)`0letle2s`, (ii) `4sletle8s`

Answer 1

Correct Answer - A::B::C::D
(a) a=slope of v-t graph
(b) `s=r_f-r_i`=net area of v-t graph
`:. r_f=r_i`+net area
`=10+10+20+10-10-10`
`=30m`
(c) (i) For `0letle2s`
u=initial velocity=0
`s_0`=initial displacement =10m
a=slope of v-t graph =`+5m//s^2`
`:. s_0+ut+1/2at^2=10+2.5t^2`
(ii) For `4sletle8s`
`u`=initial velocity =`10 m//s`
=velocity at 4s
`s_0=(10m)`+area of v-t graph upto 4s
`=10+10+20=40m`
`a`=slope of v-t graph =`-5 m//s^2`
`s=s_0+u(t-4)+1/2a(t-4)^2`
`=40+10(t-4)-2.5(t-4)^2`