A bus starts moving with 1 ms^(-2) acceleration. A man 48 m away from bus starts moving with 10 ms to catch the bus, then man will reach to bus after ........

A bus starts moving with 1 ms^(-2) acceleration. A man 48 m away from

1.	A bus starts moving with 1 ms^(-2) acceleration. A man 48 m away from bus starts moving with 10 ms to catch the bus, then man will reach to bus after ........
Answer» 5 seconds 6 seconds 7 seconds 8 seconds Solution :Suppose, ,MAN reaches to bus at distance .d.. Equation of motion for bus, In `d = v _(0)t + 1/2 at ^(2) , v _(0) =0` `d = 1/2 at ^(2) ""…(1)` and distance covered by man, `48 +d =UT` `d = 10 t -48""…(2)` `therefore` From equation (1) and (2), `1/2 at^(2) = 10t -48` `1 xx t ^(2) =20t -96""[a=1]` `t ^(2) - 20 t + 96 =0` `therefore (t -12) (t-8) =0` `therefore t =12 s or t =8s` But 12 s is not GIVEN in options. Thus, answer `t=8s`