A particle is projected vertically upwards with velocity `40m//s.` Find the displacement and distance travelled by the particle in
(a) `2s` (b) `4s` (c) `6s` Take `g=10 m//s^2`

Answer

A

Explanation

Here, u is positive (upwards) and a is negative (downwards). So, first we will find
`t_0`, the time when velocitybecomes zero.
`t_0=|u/a|=40/10=4s`
`(a) t lt t_0.` Therefore, distance and displacement are equal.
`d=s=ut+1/2at^2=40xx2-1/2xx10xx4=60m`
(b) `t=t_0.` So, again distance and displacement are equal.
`d=s=40xx4-1/2xx10xx16=80m`
(c) `t gt t_0.` Hence, `d gt s`, `s=40xx6-1/2xx10xx36=60m`
While `d=|u^2/(2a)|+1/2|a(t-t_0)^2|`
`=((40^2))/(2xx10 )+1/2xx10xx(6-4)^2`
`= 100m`