TS EAMCET 5 May 2018 Shift 1 Solved Paper

The variation of velocity along a circular path is given as,
v=2t
rω=2t
r

dθ

dt

=2t
Integrating both sides we get,
r‌

4π

∫

0

dθ=2‌

t

∫

0

tdt

t2

r

=4π
The expression for centripetal acceleration is given as,
ac=

v2

r

ac=

(2t)2

r

ac=4

t2

r

ac=4×4π
ac=16πm∕s2
The expression for tangential acceleration is given as,
aT=

dv

dt

aT=

d(2t)

dt

aT=2m∕s2
The net acceleration is given by
anet=√aT2+ac2
anet=√22+(16π)2
anet=2√1+64π2