Question

A particle starts from origin at t=0 with velocity `5.0hati m//s` and move in x.y plane under action of a force which produces a constant acceleration of `(3.0hati+2.0hatj) m//s^(2)`. (a) What is the y-coordinate of the particle at the instant its x-ccordinate is 84 m? (b) What is the speed of the particle at this time?

Answer 1

(a) `u=5.0hati m//s,a=(3.0hati+2.0hatj)m//s^(2)`
at time t, x-coordinate of particle `x=u_(x)t+(1)/(2)a_(x)t^(2)rArr84=5t+(1)/(2)xx3t^(2)+10t-168=0`
`rArr3t^(2)+28t-168=0rArrt=6s,-(28)/(3)s`, so time t=6s
y coordinate at t=6s y`=u_(y)t+(1)/(2)a_(y)t^(2)=0+(1)/(2)xx2xx(6)^(2)=36m`
(b) velocity of particle at time `t vecv=(u_(x)+a_(x)t)hati+(u_(y)+a_(y)t)hatj=(5+3xx6)hati+(0+2xx6)hatj=(23hati+12hatj)`
speed `=sqrt((23)^(2)+(2)^(2))=25.94 ms^(-1)`