An object is moving along a straight line path with a uniform accleration of ` 4 ms^"(-2)`. Intitially its lociation is at ` 5 ms` and velcity is ` 3 ms^(-1)`. ltbRgt (i) What will be position and the velocity of object at time ` t=3 s?
(ii) What will be the position of the object when it has attained a velcotiy` 8 ms^(-1)` ?

Explanation

Here, ` a= 4 ms^(-2)` , ltBrgt At , ` t=0, s(0) = m , u= 3 ms^(-1)`
What ` t= 3 s, velocity of the object is given by
` v= u + at = 3 + 4 + 3 = 15 ms^(-1)`
Position of the object at 1 t= 3 s` is given by
` s= s_0 + ut + 1/2^2`
` = 5 + 3 xx 3 = 1/2 xx 4 3of object is ` 8 ms^(-1)` is given by , i.e. ltbRgt `v^2 - u"^2 = 2 a (s-ts_0)`
or ` s_t = ( v^2 -u^2) /( 2 a) + s_0 = ( 8^2 -3^2)/( 2 xx 4) + 5`
`= (55) / 8 + 5 = 11. 88 m` .