AP EAMCET Engineering 22 Sep 2020 Shift 1 Paper

If a be the acceleration of the body, then distance travelled by the body in 2s is given as.
xi=ut+‌

1

2

at2=0×2+‌

1

2

a×(2)2
x1=2a....(i)
Velocity of body at the end of 2s is given as.
v=u+af=0+a×2
v=2a ....(ii)
Distance travelled by the body in next 2 s is given as.
x2=vt+‌

1

2

at2
=2a×2+‌

1

2

a×(2)2[ from Eq. (ii) ]
=4a+2a
x2=6a=3×2a
⇒‌‌x2=3×x1 [from Eq. (i)]
⇒‌‌x2=3x1