TS EAMCET 5 May 2018 Shift 2 Solved Paper

According to question,
The distance travelled in first 1 s is calculated as,
s1=1m=

1

2

at2
The velocity after 1 s is calculated as,
vi=u+at
=0+2×1ms−1
=2ms−1
At the end of last second, let velocity is vms−1.
Then from,
s=ut+

1

2

at2
39=v×1+

1

2

(2)×12
vf=38ms−1
Calculate the acceleration.
a=

vf−vi

t

t=

vf−vi

a

=

38−2

2

=18s
Calculate the distance covered in these 18s
s=ut+

1

2

at2
=2×18+

1

2

(2)×182
=36+324
=360m
Therefore, the distance between the points A and B is
DAB=1+39+360m
=400m