Question

Distance travelled in ( 5^{text {th }} ) half gec
( = ) Dist ( ^{circ} ) travelled in ( 2 cdot 5 ) fec -
Dist. travelled in 2 sec
( Rightarrow quad 0.5 aleft(t_{1}^{2}-t_{2}^{2}right) )
[
begin{array}{l}
=0.5 times 2left(2.5^{2}-2^{2}right)
=1.0(6.25-4)
=1(2.25)=frac{2.25 mathrm{m}}{.}
end{array}
]

# 18. A particle starts moving with acceleration 2 m/s2. Distance travelled h (1) 1.25 m (2) 2.25 m (3) 6.25 m Solution of Assignment (Set-2) eleration 2 m/s2. Distance travelled by it in 5th half second is (4) 30.25 m

Solution