A particle moves along the x-direction with constant acceleration. The displacement, measured from a convenient position, is 2 m at time `t= 0` and is zero when `t= 10 s.` If the velocity of the particle is momentary zero when `t = 6 s,` determine the acceleration a and the velocity v when `t= 10s.`

Answer

A, B

Explanation

`s=s_0+ut+1/2at^2`
at `t=0, s=s_0=2m …(i)`
at `t=10s,`
`s=0=s_0+ut+1/2at^2`
or `10u+50a=-2….(ii)`
`v=u+at`
`:. 0=u+at`
`:. 0=u+6a (at t=6s)`
`:. u+6a=0….(iii)`
Solving Eqs. (ii) and (iii) we get,
`u=-1.2 m//s`
and `a=0.2 m//s^2`
Now, `v=u+at`
`:. v=(-12)+(0.2)(10)`
`=0.8 m//s`