Question

A particle starts from rest and moves with a constant acceleration of `4 m//s^(2)` for `10 s`, then it moves under constant seceleration and stops in `20 s`. Find (a) the maximum velocity attained by the particle (b) the total distance traveled and (c ) the distance/distances from the origin, when the particle is moving at half the maximum velocity.

Answer 1

`A` to `B v=u+at`
`v_(m)=0+a_(1)t_(1)=4xx10=40 m//s` (maximum velocity )
`d_(1)=1/2a_(1)t_(1)^(2)=1/2xx4xx(10)^(2)=200m`
`B` to `C: v=u+at`
`0=v_(m)-a_(2)t_(2)=40-a_(2)xx20`
`impliesa_(2)=m//s^(2)` (deceleration)
`d_(2)=v_(m)t_(2)-1/2 a_(2)t_(2)^(2)=40xx20-1/2xx2xx(20)^(2)`
`=800-400=400m`
Total distance traveled `=d_(1)+d_(2)=800m`
(c )
Distance//distances at which particle velocity=`v_(m)//2`
`A` to `D`: `(v_(m)/2)^(2)=0+2a_(1)s_(1)`
`implies (20)^(2)=2xx4xxs_(1)impliess_(1)=50m`
`E` to `C`: `v^(2)=(v_(m)/2)^(2)-2a_(2)s_(0)`
`implies0=(20)^(2)-2xx2s_(0)impliess_(0)=100m`
Distance `s_(2)=(d_(1)+d_(2))-s_(0)=800-100=700m`
Distance/distances from origin `A: 50 m, 700 m`