Question

A particle starts with an initial velocity 2.0 m/s along the positive x-direction and it accelerations uniformly at the rate of `0.40m//s^(2)`
(a) Find the distance travelled by it in the first three seconds.
(b) How much time does it take to reach the velocity 9.0 m/s?
(c) How much distance will it cover in reaching the velocity 9.0 m/s?

Answer 1

Given that `y=2m//s, a=0.40m//s^(2)`
(a) We have to calculate s when t = 3 s
`s=ut+(1)/(2)at^(2)=3xx2+(1)/(2)xx0.40xx3^(2)`
`=6+0.20xx9=6+1.80=7.80m`
(b) We have to calculate t, when v = 9 m/s
As `v=u+at,` So, `t=(v-u)/(a)=(9-2)/(0.4)=(70)/(4)=(35)/(2)=17.5s`
(c) Here, s = ?, v = 9 `m//s^(2)`, As `v^(2)=u^(2)+2as`
So, `s=(v^(2)-u^(2))/(2a)=(9^(2)-(2)&(2))/(2xx0.4)=(81-4)/(0.8)=(77)/(0.8)=(770)/(8)=(385)/(4)m`